(1) The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 94 inches and a standard deviation of 14 inches. What is the probability that the mean annual snowfall during 49 randomly picked years will exceed 96.8 inches? 0.0808 0.5808 0.0026 0.4192 (2) The scores on a certain test are normally distributed with a mean score of 43 and a standard deviation of 3. What is the probability that a sample of 90 students will have a mean score of at least 43.3162? 0.3174 0.8413 0.1587 0.3413 (3) The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability of more than 53 correct answers The area to the right of 53.5 The area to the left of 53.5 The area to the right of 53 (4) A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 7.7 hours. The area to the right of 52.5 0.9634 0.9931 0.9712 0.8531 (5) The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability of at least 48 boys The area to the left of 47.5 The area to the right of 47.5 The area to the right of 48 The area to the right of 48.5 (6) The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability of exactly 47 green marbles The area between 46.5 and 47.5 The area between 46.5 and 47 The area between 46.5 and 48.5 The area between 47 and 47.5 (7) The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability that the number of correct answers is between 18 and 45 inclusive The area between 17.5 and 44.5 The area between 17.5 and 45.5 The area between 18.5 and 44.5 The area between 18 and 45 (8) For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n = 22 and p = 0.6 Normal approximation is suitable. Normal approximation is not suitable. (9) For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n = 19 and p = 0.8 Normal approximation is suitable. Normal approximation is not suitable. (10) Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. With n = 20 and p = 0.60, estimate P(fewer than 8). 0.0202 0.4953 0.0668 0.4332 (11) Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. A multiple choice test consists of 60 questions. Each question has 4 possible answers of which one is correct. If all answers are random guesses, estimate the probability of getting at least 20% correct. 0.0901 0.8508 0.1492 0.3508 (12) Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. A certain question on a test is answered correctly by 22% of the respondents. Estimate the probability that among the next 150 responses there will be at most 40 correct answers. 0.8997 0.0694 0.9306 0.1003 (13) All of the following are considered to be unbiased estimators of a statistic except: Range Variance Mean Proportion (14) All of the following are considered to be biased statistical estimators except: Range Median Proportion Standard Deviation (15) In very large samples, which biased statistical estimator is sometimes used because the bias is relatively small? Mean Variance Range Standard Deviation (16) Which two rules must be considered in order to apply the Central Limit Theorem to a data set? The sample size and distribution The mean and median The standard error and sample size The distribution and range (17) Which of the following distributions will approach a normal distribution as the sample size increases? U-Shape Normal All of these will approach normal distribution as the sample size increases. Uniform (18) If, under a given assumption, the probability of a particular observed event is exceptionally small (such as less than 0.05), we can conclude that the assumption is probably not correct. This defines the following: Central Limit Theorem Rare Event Rule for Inferential Statistics Continuity Correction Range Rule of Thumb (19) All of the following are steps in the procedure for using normal distribution to approximate a binomial distribution except: Finding the appropriate z-score and using your calculator to find the area (probability). Verify that np and nq are both exactly 5. Identify and adjust the discreet whole number being used by the continuity correction. Find the values for the mean and standard deviations using special formulas. (20) The notation for the sampling distribution of x shows an adjustment for the sample means because: the formula needs to take the sample size into account. As the sample size increases, the distribution spreads to uniform. Biased estimators cause several adjustments. The distribution is never normal. (1) Hypothesis testing includes all of the following concepts except: identifying a P-value. identifying a null and alternative hypothesis. voluntary response samples and biased funding sources. identifying a critical value. (2) True or False: The null hypothesis contains the equality. True False (3) Test statistics can be in any of the following forms except: X values 2 Z-scores Student t values P-values (4) The Pvalue is a probability value. It is an area (to either side) of the: evidence alpha critical value test statistic (5) What specific wording would you use if your original claim does not contain an equality and you do reject the null? There is not suffient sample evidence to support the claim that ...(original claim). There is sufficient evidence to warrent the rejection of the claim that ... (original claim). The sample data support the claim that ... (original claim). There is not sufficiant evidence to warrent rejection of the claim that ...(original claim) (6) What specific wording would you use if your original claim does contain an equality and you do reject the null? There is sufficient evidence to warrent the rejection of the claim that ... (original claim). There is not sufficiant evidence to warrent rejection of the claim that ...(original claim). There is not suffient sample evidence to support the claim that ...(original claim). The sample data support the claim that ... (original claim). (7) Suppose that you wish to test a claim about a population mean. Which distribution should be used given that the sample is a simple random sample, is unknown, n = 15, and the population is not normally distributed? t-distribution X 2 normal (z-score) distribution Neither the normal (z-scores) nor the t-distribution (8) Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol ( p or ) for the indicated parameter. An entomologist writes an article in a scientific journal which claims that fewer than 7 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light. H : p > 0.0007 o H : p 0.0007 H : p < 0.0007 1 o H : p 0.0007 H : p = 0.0007 1 o H : p > 0.0007 1 (9) Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol ( p or ) for the indicated parameter. Carter Motor Company claims that its new sedan, the Libra, will average better than 26 miles per gallon in the city. Use , the true average mileage of the Libra. H: o = 26 H : > 26 H : = 26 1 o H : < 26 H : < 26 1 o H: 1 26 (10) Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.05 for a twotailed test. 1.764 1.645 1.96 2.575 (11)Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.09 for a righttailed test 1.96 1.96 1.34 1.34 (12)Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.05 for a lefttailed test. 1.645 -1.96 -1.645 (13) Find the value of the test statistic z using the test statistic for a proportion conversion formula found on p. 397. A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1671 subjects with 30% saying that they play a sport. -16.35 16.35 33.38 -33.38 (14)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a righttailed test is z = 0.52. 0.6030; fail to reject the null hypothesis 0.3015; reject the null hypothesis 0.3015; fail to reject the null hypothesis 0.0195; reject the null hypothesis (15)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a lefttailed test is z = 1.83. 0.0672; reject the null hypothesis 0.9664; fail to reject the null hypothesis 0.0336; reject the null hypothesis 0.0672; fail to reject the null hypothesis (16)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a twotailed test is z = 1.95. 0.0512; fail to reject the null hypothesis 0.0512; reject the null hypothesis 0.0256; reject the null hypothesis 0.9744; fail to reject the null hypothesis (17) Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. Use pp. 404 405 for support. A medical researcher claims that 6% of children suffer from a certain disorder. Identify the type I error for the test. Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually different from 6%. Reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%. Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%. Reject the claim that the percentage of children who suffer from the disorder is different from 6% when that percentage really is different from 6%. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or (18) type II error for the test. Use pp. 404 405 for support. The principal of a school claims that the percentage of students at his school that come from single parent homes is 13%. Identify the type II error for the test. Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually 13%. Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually different from 13% Reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually 13%. Reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually less than 13%. (19) Review p. 406. In conducting hypothesis tests, you may use the Pvalue method and compare the Pvalue based on the critical value with the alpha area. You can use the traditional method by graphing the critical value and determining if it lies in the critical (red) region or not. You can also construct a confidence interval like we did in Chapter 7. A confidence interval estimate contains: the likely values of that paramater Type I and Type II errors 95% of the data the original claim (20) Using the traditional method of hypothesis testing, if the test statistic falls in the critical (red) region, you should: reject the null fail to reject the null accept the null start over (1) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : = 22; H : 22 o 1 H : 22; H : = 22 o 1 H : = 22; H : < 22 o 1 H : p = 22; H : p 22 o 1 (2) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = 1.52 Test statistic: z = 0.005 Test statistic: z = 10.33 Test statistic: z = -10.33 Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. (3) Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 1 and p. 400. P-value: 0.008 P-value: 0.005 P-value: 0.0002 P-value: 1.645 Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. (4) Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (5) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Use the flow chart on p. 403 to select the final wording for your answer. There is not sufficient evidence to warrant rejection of the claim that the population mean temperature is 22 degrees Celcius. There is sufficient evidence to warrant rejection of the claim that the population mean temperature is 22 degrees Celcius. The sample data support the claim that the population mean temperature is 22 degrees Celcius. There is not sufficientsample evidence to support the claim that the population mean temperature is 22 degrees Celcius. (6) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : < 200; H : > 200 o 1 H : = 200; H : < 200 o 1 H : = 200; H : o 1 200 H : < 200; H : = 200 o 1 (7) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = -2.33 Test statistic: z = 0.98 Test statistic: z = -0.98 Test statistic: z = 0.10 (8) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 6 and p. 400. P-value: 0.100 P-value: 0.1635 P-value: 0.90 P-value: 1.645 (9) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (10)The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Use the flow chart on p. 403 to select the final wording for your answer. There is not sufficient sample evidence to support the claim that the mean is less than 200 pounds. The sample data support the claim that the mean is less than 200 pounds. There is not sufficient evidence to warrant rejection of the claim that the mean is less than 200 pounds. There is sufficient evidence to warrant rejection of the claim that the mean is less than 200 pounds. (11)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : = 39.9; H : 39.9 o H : o 1 39.9; H : = 39.9 1 H : = 39.9; H : > 39.9 o 1 H : = 40.5; H : 40.5 o 1 (12) A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = 3.75 Test statistic: z = 3.21 Test statistic: z = 1.645 Test statistic: z = -3.75 (13)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 11 and p. 400. P-value: 0.05 P-value: 0.005 P-value: 0.0002 P-value: 0.02 (14)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (15)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Use the flow chart on p. 403 to select the final wording for your answer. There is sufficient evidence to warrant rejection of the claim that the mean equals 39.9 cm. There is not sufficient sample evidence to support the claim that the population mean equals 39.9 cm. There is not sufficient evidence to warrant rejection of the claim that the population mean equals 39.9 cm. . The sample data support the claim that the population mean equals 39.9 cm. (16)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 981. Sample data: n = 24, sample mean = 972, s = 26. The sample data appear to come from a normally distributed population with = 28. Normal Student t Neither (17)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 120. Sample data: n = 11, sample mean = 100, s = 15.2. The sample data appear to come from a normally distributed population with unknown and . Normal Neither Student t (18)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 82. Sample data: n = 20, sample mean = 100, s = 15.1. The sample data appear to come from a population with a distribution that is very far from normal, and is unknown. Normal Neither Student t (19)A simple random sample of 40 recorded speeds (in mi/ hr) is obtained from cars traveling on a section of Highway 405 in Los Angeles. The sample has a mean of 68.4 mi/ h and a sample standard deviation of 5.7 mi/h (based on data from Sigalert). Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. Ho : > 65 mi/h; H1 : = 65 mi/h H : = 65 mi/h; H : o 1 65 mi/h H : = 65 mi/h; H : > 65 mi/h o 1 H : = 65 mi/h; H : < 65 mi/h o 1 (20)A simple random sample of 40 recorded speeds (in mi/ hr) is obtained from cars traveling on a section of Highway 405 in Los Angeles. The sample has a mean of 68.4 mi/ h and a sample standard deviation of 5.7 mi/h (based on data from Sigalert). Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h. Use Step 6 on p. 434 to find the test statistic student t value. (Be sure to record this for future use.) t = 2.35 t = 3.773 t = .005 t = 1.685 (1) Hypothesis testing includes all of the following concepts except: identifying a P-value. identifying a null and alternative hypothesis. voluntary response samples and biased funding sources. identifying a critical value. (2) True or False: The null hypothesis contains the equality. True False (3) Test statistics can be in any of the following forms except: X values 2 Z-scores Student t values P-values (4) The Pvalue is a probability value. It is an area (to either side) of the: evidence alpha critical value test statistic (5) What specific wording would you use if your original claim does not contain an equality and you do reject the null? There is not suffient sample evidence to support the claim that ...(original claim). There is sufficient evidence to warrent the rejection of the claim that ... (original claim). The sample data support the claim that ... (original claim). There is not sufficiant evidence to warrent rejection of the claim that ...(original claim) (6) What specific wording would you use if your original claim does contain an equality and you do reject the null? There is sufficient evidence to warrent the rejection of the claim that ... (original claim). There is not sufficiant evidence to warrent rejection of the claim that ...(original claim). There is not suffient sample evidence to support the claim that ...(original claim). The sample data support the claim that ... (original claim). (7) Suppose that you wish to test a claim about a population mean. Which distribution should be used given that the sample is a simple random sample, is unknown, n = 15, and the population is not normally distributed? t-distribution X 2 normal (z-score) distribution Neither the normal (z-scores) nor the t-distribution (8) Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol ( p or ) for the indicated parameter. An entomologist writes an article in a scientific journal which claims that fewer than 7 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light. H : p > 0.0007 o H : p 0.0007 H : p < 0.0007 1 o H : p 0.0007 H : p = 0.0007 1 o H : p > 0.0007 1 (9) Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol ( p or ) for the indicated parameter. Carter Motor Company claims that its new sedan, the Libra, will average better than 26 miles per gallon in the city. Use , the true average mileage of the Libra. H: o = 26 H : > 26 H : = 26 1 o H : < 26 H : < 26 1 o H: 1 26 (10) Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.05 for a twotailed test. 1.764 1.645 1.96 2.575 (11)Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.09 for a righttailed test 1.96 1.96 1.34 1.34 (12)Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.05 for a lefttailed test. 1.645 -1.96 -1.645 (13) Find the value of the test statistic z using the test statistic for a proportion conversion formula found on p. 397. A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1671 subjects with 30% saying that they play a sport. -16.35 16.35 33.38 -33.38 (14)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a righttailed test is z = 0.52. 0.6030; fail to reject the null hypothesis 0.3015; reject the null hypothesis 0.3015; fail to reject the null hypothesis 0.0195; reject the null hypothesis (15)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a lefttailed test is z = 1.83. 0.0672; reject the null hypothesis 0.9664; fail to reject the null hypothesis 0.0336; reject the null hypothesis 0.0672; fail to reject the null hypothesis (16)Use the given information to find the Pvalue. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a twotailed test is z = 1.95. 0.0512; fail to reject the null hypothesis 0.0512; reject the null hypothesis 0.0256; reject the null hypothesis 0.9744; fail to reject the null hypothesis (17) Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. Use pp. 404 405 for support. A medical researcher claims that 6% of children suffer from a certain disorder. Identify the type I error for the test. Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually different from 6%. Reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%. Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%. Reject the claim that the percentage of children who suffer from the disorder is different from 6% when that percentage really is different from 6%. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or (18) type II error for the test. Use pp. 404 405 for support. The principal of a school claims that the percentage of students at his school that come from single parent homes is 13%. Identify the type II error for the test. Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually 13%. Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually different from 13% Reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually 13%. Reject the claim that the percentage of students that come from single-parent homes is equal to 13% when that percentage is actually less than 13%. (19) Review p. 406. In conducting hypothesis tests, you may use the Pvalue method and compare the Pvalue based on the critical value with the alpha area. You can use the traditional method by graphing the critical value and determining if it lies in the critical (red) region or not. You can also construct a confidence interval like we did in Chapter 7. A confidence interval estimate contains: the likely values of that paramater Type I and Type II errors 95% of the data the original claim (20) Using the traditional method of hypothesis testing, if the test statistic falls in the critical (red) region, you should: reject the null fail to reject the null accept the null start over (1) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : = 22; H : 22 o 1 H : 22; H : = 22 o 1 H : = 22; H : < 22 o 1 H : p = 22; H : p 22 o 1 (2) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = 1.52 Test statistic: z = 0.005 Test statistic: z = 10.33 Test statistic: z = -10.33 Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. (3) Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 1 and p. 400. P-value: 0.008 P-value: 0.005 P-value: 0.0002 P-value: 1.645 Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. (4) Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (5) Various temperature measurements are recorded at different times for a particular city. The mean of 20C is obtained for 60 temperatures. Assuming that : = 1.5, test the claim that the population mean is 22C . Use a 0.05 significance level. Use the flow chart on p. 403 to select the final wording for your answer. There is not sufficient evidence to warrant rejection of the claim that the population mean temperature is 22 degrees Celcius. There is sufficient evidence to warrant rejection of the claim that the population mean temperature is 22 degrees Celcius. The sample data support the claim that the population mean temperature is 22 degrees Celcius. There is not sufficientsample evidence to support the claim that the population mean temperature is 22 degrees Celcius. (6) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : < 200; H : > 200 o 1 H : = 200; H : < 200 o 1 H : = 200; H : o 1 200 H : < 200; H : = 200 o 1 (7) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = -2.33 Test statistic: z = 0.98 Test statistic: z = -0.98 Test statistic: z = 0.10 (8) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 6 and p. 400. P-value: 0.100 P-value: 0.1635 P-value: 0.90 P-value: 1.645 (9) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (10)The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Use the flow chart on p. 403 to select the final wording for your answer. There is not sufficient sample evidence to support the claim that the mean is less than 200 pounds. The sample data support the claim that the mean is less than 200 pounds. There is not sufficient evidence to warrant rejection of the claim that the mean is less than 200 pounds. There is sufficient evidence to warrant rejection of the claim that the mean is less than 200 pounds. (11)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. H : = 39.9; H : 39.9 o H : o 1 39.9; H : = 39.9 1 H : = 39.9; H : > 39.9 o 1 H : = 40.5; H : 40.5 o 1 (12) A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Use Step 6 on p. 427 to find the test statistic zscore. (Be sure to record this for future use.) Test statistic: z = 3.75 Test statistic: z = 3.21 Test statistic: z = 1.645 Test statistic: z = -3.75 (13)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Find the Pvalue of the test statistic zscore. Use what you know about the 'tails' identified from your work in question 11 and p. 400. P-value: 0.05 P-value: 0.005 P-value: 0.0002 P-value: 0.02 (14)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Compare the Pvalue to the alpha significance level. Compare the test statistic from question 2 with the critical value based off of the given significance level. Will you reject the null or fail to reject the null hypothesis? Reject the null hypothesis Fail to reject the null hypothesis (15)A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. Use the flow chart on p. 403 to select the final wording for your answer. There is sufficient evidence to warrant rejection of the claim that the mean equals 39.9 cm. There is not sufficient sample evidence to support the claim that the population mean equals 39.9 cm. There is not sufficient evidence to warrant rejection of the claim that the population mean equals 39.9 cm. . The sample data support the claim that the population mean equals 39.9 cm. (16)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 981. Sample data: n = 24, sample mean = 972, s = 26. The sample data appear to come from a normally distributed population with = 28. Normal Student t Neither (17)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 120. Sample data: n = 11, sample mean = 100, s = 15.2. The sample data appear to come from a normally distributed population with unknown and . Normal Neither Student t (18)Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither. Claim: = 82. Sample data: n = 20, sample mean = 100, s = 15.1. The sample data appear to come from a population with a distribution that is very far from normal, and is unknown. Normal Neither Student t (19)A simple random sample of 40 recorded speeds (in mi/ hr) is obtained from cars traveling on a section of Highway 405 in Los Angeles. The sample has a mean of 68.4 mi/ h and a sample standard deviation of 5.7 mi/h (based on data from Sigalert). Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h. Identify the null and alternative hypothesis. Record what kind of test (re: tails) you will be conducting for your notes. Ho : > 65 mi/h; H1 : = 65 mi/h H : = 65 mi/h; H : o 1 65 mi/h H : = 65 mi/h; H : > 65 mi/h o 1 H : = 65 mi/h; H : < 65 mi/h o 1 (20)A simple random sample of 40 recorded speeds (in mi/ hr) is obtained from cars traveling on a section of Highway 405 in Los Angeles. The sample has a mean of 68.4 mi/ h and a sample standard deviation of 5.7 mi/h (based on data from Sigalert). Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mi/h. Use Step 6 on p. 434 to find the test statistic student t value. (Be sure to record this for future use.) t = 2.35 t = 3.773 t = .005 t = 1.685