SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. 1) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 11 8 8 10 11 9 9 14 8 5 Find a 95% confidence interval for the population standard deviation . Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 2) A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that no more than 1% are defective. Solve the problem. 3) A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to exaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition? His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 4) 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. Test the given claim. Use the P-value method or the traditional method as indicated. Identify the null hypothesis, alternative hypothesis, test statistic, critical value(s) or P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 5))A simple random sample of 15-year old boys from one city is obtained and their weights (in pounds) are listed below. Use a 0.01 significance level to test the claim that these sample weights come from a population with a mean equal to 149 lb. Assume that the standard deviation of the weights of all 15-year old boys in the city is known to be 16.2 lb. Use the traditional method of testing hypotheses. 147 138 162 151 134 189 157 144 175 127 164 Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 6) A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below. 14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2 Test the claim at the 0.01 significance level. Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion. 7) Test the claim that the mean lifetime of car engines of a particular type is greater than 220,000 miles. Sample data are summarized as n = 23, x = 226,450 miles, and s = 11,500 miles. Use a significance level of = 0.01. 2 Find the critical value or values of x^2 based on the given information. 8) H1: < 0.14 n = 23 = 0.10 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. 9) = 0.08; H1 is 3.24 Solve the problem. 10) A simple random sample of students is selected, and the students are asked how much time they spent preparing for a test. The times (in hours) are as follows: 1.3 7.2 4.2 12.5 6.6 2.5 5.5 Based on these results, a confidence interval for the population mean is found to be = 5.7 4.4. Find the degree of confidence. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. 1) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 11 8 8 10 11 9 9 14 8 5 Find a 95% confidence interval for the population standard deviation Answer Mean,m=9.3 and standard deviation,d=2.4 Critical z-value =1.96 for 95% confidence interval. Lower limit =>m-d*z =>9.3-2.4*1.96 =4.596 Upper limit=>m+d*z =>9.3+2.4*1.96 =>14.004 Confidence interval is [4.6,14] =>Rounded to nearest tenth Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 2) A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that no more than 1% are defective. Answer Null Hypothesis:Percentage defective is,p<= .01 Alternative Hypothesis:Percentage defective p>.01 Test statistic=z-test for proportion We have p-hat=.03 Standard deviation =>sqrt(.01*.03/600)=.00406 Zscore => (.03-.01)/.00406 =4.926 P value is 4.25*10^-7 As the p-value is less than .01 threfore we reject that no more than 1% of digital memory cards are defective. Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.01 Level of Significance 0.01 Number of Items of Interest 18 Sample Size 600 Intermediate Calculations Sample Proportion Standard Error Z Test Statistic Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 0.03 0.004 1 4.923 7 2.326 3 0.000 0 Solve the problem. 3) A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to exaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition? His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed. Answer Null Hypothesis: p=.984 Alternative Hypothesis:p<.984 We have, p=.04 Standard error:sqrt(.05*.95/100)=.02179 Confidence interval at .05 significance level is => .04 +/- 1.645*.02179 =>[.0157,.0984] Highest rate=>1-.0157=98.4% (answer) Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 4) 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. Test the given claim. Use the P-value method or the traditional method as indicated. Identify the null hypothesis, alternative hypothesis, test statistic, critical value(s) or P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Answer Null Hypothesis:m=200 lb Alternative Hypothesis:m <200 lb here ,n=54,standard deviation,d=121.2 m=183.9 z score=(183.9-200)/(121.2/sqrt54 )=-16.1/16.493 >.1 therefore accept the null hypothesis and reject the claim that the mean is less than 200 lb. 5))A simple random sample of 15-year old boys from one city is obtained and their weights (in pounds) are listed below. Use a 0.01 significance level to test the claim that these sample weights come from a population with a mean equal to 149 lb. Assume that the standard deviation of the weights of all 15-year old boys in the city is known to be 16.2 lb. Use the traditional method of testing hypotheses. 147 138 162 151 134 189 157 144 175 127 164 Answer Null Hypothesis:m=149 Alternative Hypothesis:m=/= 149 Zstatistics =(153.45-149)/16.2/sqrt11 =.91 The p-value is .8185 As the p-value is greater than .01 therefore we accept the null hypothesis. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 6) A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below. 14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2 Test the claim at the 0.01 significance level. Answer Mean of the sample =14.05 Standard deviation of the sample =.324 t-test => Null Hypothesis:Mean weight=14 oz Alternative hypothesis;Mean weight =/= 14 oz. t-statistic =(14.05-14)/.324/sqrt8 = .436 Critical t-value=+- 3.499. Thus it fails to reject H0. Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion. 7) Test the claim that the mean lifetime of car engines of a particular type is greater than 220,000 miles. Sample data are summarized as n = 23, x = 226,450 miles, and s = 11,500 miles. Use a significance level of = 0.01. Answer test statistics =(226,450-220,000)/(11,500/sqrt23)=2.6898 p value of the test statistics =>.0066 As the p value is greater than .01 therefore we reject the null hypothesis. Thus, the mean lifetime of carengine is greater than 220,000 miles. 2 Find the critical value or values of x^2 based on the given information. 8) H1: < 0.14 n = 23 = 0.10 Answer Critical value=14.042 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. 9) = 0.08; H1 is 3.24 Answer As it is a two tailed test therefore the critical value is (+/- 1.75)=>answer Solve the problem. 10) A simple random sample of students is selected, and the students are asked how much time they spent preparing for a test. The times (in hours) are as follows: 1.3 7.2 4.2 12.5 6.6 2.5 5.5 Based on these results, a confidence interval for the population mean is found to be = 5.7 4.4. Find the degree of confidence. Answer Degree of confidence can be calculated by first calculating the critical z value We have,standard error =4.4 MOE =t*SE SE=standard error =standard deviation/sqrtn =3.6785/sqrt7 =1.39 Thus 4.4=z*1.39 t=3.165 From the two-tailed test with degree of freedom=6 and t=3.165 is .02 Therefore the confidence level is 98%