Course #: MATH 2209 Course Name: Introduction to Statistics II Module One: The CLT, t-distributions and Inference for the Mean Assignment 1 Name: ID: 1. The weight of Skittles in a small bag is stated to be 61.5 g. The amount that the packaging machine puts in these bags is believed to have a Normal model with mean 65 g and standard deviation 1.8 g. a. What fraction of all bags sold are underweight? b. Some of the packages of Skittles are sold in \"bargain packs\" of three bags. What is the probability that none of the three is underweight? c. What is the probability that the mean weight of the three bags is below the stated amount? d. What is the probability that the mean weight per bag of a 24-bag case of Skittles is below 65 g? 2. Receipts from a drug store show that customer purchases have a skewed distribution with a mean of $32 and a standard deviation of $20. a. Explain why you cannot determine the probability that the next customer will spend at least $40. b. Can you estimate the probability that the next 10 customers will spend an average of at least $40? Explain. c. Is it likely that the next 50 customers will spend an average of at least $40? Explain. 3. A random sample of size ten was drawn from a population of 965 systolic blood pressure rates for males aged 15-20 years. The Minitab output showing the summary statistics for this sample of 10 is given below. Descriptive Statistics: C2 Variable C2 N 10 N* 0 Variable C2 Maximum 140.00 Mean 117.10 SE Mean 4.90 StDev 15.48 Minimum 90.00 Q1 104.00 Median 120.00 Q3 128.00 a. Identify the population and the sample. b. Use the sample to calculate a 95% confidence interval for the average systolic blood pressure rate of all males. c. Interpret this interval in the context of the problem. d. At the 5% level of significance, does the data provide evidence that the mean systolic blood pressure rate is different from 120? i. State the hypotheses to be tested, and define the parameter. ii. Check the required assumptions in the context of the question. iii. Conduct the test and find the P-value iv. State your conclusion in the context of the question. Course #: MATH 2209 Course Name: Introduction to Statistics II Module Two: Comparing Two Means Assignment 2 Name: ID: 1. In an effort to improve final grades in her class, a professor tried a new method of evaluation with one of her intro stats classes last term. A test of the null hypothesis that her old method (used with one class) and new method (used with her other class) don't differ in the mean final grade for students yielded a p-value of 0.066. Would a 95% confidence interval for new - old include 0? Explain. 2. Students at a very large university have the option to take an introductory algebra class on campus or through distance learning. Dr. Brown teaches both the on campus and distance courses and gives all students the same final exam regardless of how they take the course. He is interested in whether there is a difference between the final exam grades depending on how the students take the course. He randomly selects 12 distance students and 12 on campus students. A Minitab output showing the data from his samples is below. Two-Sample T-Test and CI: Distance, On Campus Two-sample T for Distance vs On Campus Distance On Campus N 12 12 Mean 66.17 80.6 StDev 4.91 11.0 SE Mean 1.4 3.2 Difference = (Distance) - (On Campus) Estimate for difference: -14.42 95% CI for difference: (-21.85, -6.98) a. Interpret this interval in the context of the question. b. What does this interval suggest about the difference between the two courses? 3. Military intelligence units are often responsible for intercepting and deciphering coded messages used by enemy forces. In this type of work, time is critically important. Recently, the army compared two methods of deciphering coded messages. Two special agents randomly assigned 49 messages to two groups, and used identical computers to decode the messages. One of them used Deciphermatic (sample 1) to decode 24 messages and the other used QuickDecipher (sample 2) to decode 25 messages. The means and standard deviations for the time required by the computer, in minutes to decode the messages are given in the Minitab output below. Test an appropriate hypothesis and use the p-value to state a conclusion. Be sure to follow the full steps to hypothesis testing, in the context of the question. Two-Sample T-Test and CI Sample N Mean StDev 1 24 37.48 8.16 2 25 31.02 3.68 SE Mean 1.7 0.74 Difference = mu (1) - mu (2) Estimate for difference: 6.46 95% CI for difference: (2.75, 10.17) T-Test of difference = 0 (vs not =): T-Value = 3.55 P-Value=0.001 DF=31 Course #: MATH 2209 Course Name: Introduction to Statistics II Module One: The CLT, t-distributions and Inference for the Mean Assignment 1 Name: ID: 1. The weight of Skittles in a small bag is stated to be 61.5 g. The amount that the packaging machine puts in these bags is believed to have a Normal model with mean 65 g and standard deviation 1.8 g. a. What fraction of all bags sold are underweight? P(X<61.5)= P(Z<(61.5-65)/1.8) =P (Z<-1.9444)= 0.026 b. Some of the packages of Skittles are sold in \"bargain packs\" of three bags. What is the probability that none of the three is underweight? c. P(X>61.5) = P (Z<(61.5-65)/1.8/3) =P(Z>-3.3679) =0.9996 What is the probability that the mean weight of the three bags is below the stated amount? P(X<61.5) =1-0.9996=0.0004 d. What is the probability that the mean weight per bag of a 24-bag case of Skittles is below 65 g? P(X<65)=P(Z<65-65.5)/1.8/24)=P(X<0)=0.5 2. Receipts from a drug store show that customer purchases have a skewed distribution with a mean of $32 and a standard deviation of $20. a. Explain why you cannot determine the probability that the next customer will spend at least $40. Since the distribution is skewed, we cannot apply the probability that are gotten using properties of normal model. b. Can you estimate the probability that the next 10 customers will spend an average of at least $40? Explain. c. Is it likely that the next 50 customers will spend an average of at least $40? Explain. No, since the distribution is skewed, we need to a sample of at least 30 customers for the CLT to ensure a normal distribution for the sample means 3. A random sample of size ten was drawn from a population of 965 systolic blood pressure rates for males aged 15-20 years. The Minitab output showing the summary statistics for this sample of 10 is given below. Descriptive Statistics: C2 Variable C2 N 10 N* 0 Variable C2 Maximum 140.00 Mean 117.10 SE Mean 4.90 a. Identify the population and the sample. StDev 15.48 Minimum 90.00 Q1 104.00 Median 120.00 Q3 128.00 Course #: MATH 2209 Course Name: Introduction to Statistics II Module One: The CLT, t-distributions and Inference for the Mean b. Use the sample to calculate a 95% confidence interval for the average systolic blood pressure rate of all males. CI=x-bart*s/n CI=4.90 t*15.48/sqrt(10) (106.03, 128.17) c. Interpret this interval in the context of the problem. We are 95% confidence that the population mean is within (106.03, 128.17) d. At the 5% level of significance, does the data provide evidence that the mean systolic blood pressure rate is different from 120? i. State the hypotheses to be tested, and define the parameter. H0: =120 Ha: 120 ii. Check the required assumptions in the context of the question. Normality is met Sample size is less than 30 hence use t-distribution iii. Conduct the test and find the P-value P(X=120) T = (117.1- 120)/15.48/10) P-value=0.568 iv. State your conclusion in the context of the question. Reject H0 and conclude that the mean is different from 120 Course #: MATH 2209 Course Name: Introduction to Statistics II Module Two: Comparing Two Means Assignment 2 Name: ID: 1. In an effort to improve final grades in her class, a professor tried a new method of evaluation with one of her intro stats classes last term. A test of the null hypothesis that her old method (used with one class) and new method (used with her other class) don't differ in the mean final grade for students yielded a p-value of 0.066. Would a 95% confidence interval for new - old include 0? Explain. Yes, this is because p-value for the test is greater than 0.025 hence we expect the difference to be insignificant and thus the confidence interval will have a negative and a positive value hence including zero. 2. Students at a very large university have the option to take an introductory algebra class on campus or through distance learning. Dr. Brown teaches both the on campus and distance courses and gives all students the same final exam regardless of how they take the course. He is interested in whether there is a difference between the final exam grades depending on how the students take the course. He randomly selects 12 distance students and 12 on campus students. A Minitab output showing the data from his samples is below. Two-Sample T-Test and CI: Distance, On Campus Two-sample T for Distance vs On Campus Distance On Campus N 12 12 Mean 66.17 80.6 StDev 4.91 11.0 SE Mean 1.4 3.2 Difference = (Distance) - (On Campus) Estimate for difference: -14.42 95% CI for difference: (-21.85, -6.98) a. Interpret this interval in the context of the question. Suppose we repeated this study with different random samples for the two groups. Based on the confidence interval, we would expect 95% of the observed difference in sample means to be between (-21.85, -6.98) Also it can be interpreted as We are 95% confidence that the difference between the final exam grades depending on how the students take the course is within this interval b. What does this interval suggest about the difference between the two courses? Since 0 is not within the interval, thee is difference between the final exam grades depending on how the students take the course 3. Military intelligence units are often responsible for intercepting and deciphering coded messages used by enemy forces. In this type of work, time is critically important. Recently, the army compared two methods of deciphering coded messages. Two special agents randomly assigned 49 messages to two groups, and used identical computers to decode the messages. One of them used Deciphermatic (sample 1) to decode 24 messages and the other used QuickDecipher (sample 2) to decode 25 messages. The means and standard deviations for the time required by the computer, in minutes to decode the messages are given in the Minitab output below. Course #: MATH 2209 Course Name: Introduction to Statistics II Module Two: Comparing Two Means Test an appropriate hypothesis and use the p-value to state a conclusion. Be sure to follow the full steps to hypothesis testing, in the context of the question. Two-Sample T-Test and CI Sample N Mean StDev 1 24 37.48 8.16 2 25 31.02 3.68 SE Mean 1.7 0.74 Difference = mu (1) - mu (2) Estimate for difference: 6.46 95% CI for difference: (2.75, 10.17) T-Test of difference = 0 (vs not =): T-Value = 3.55 P-Value=0.001 DF=31 Answer Step 1: H0 1=2 Ha12 =0.05 Step 2: Test statistics=3.55 Step 3: P-value=0.001 Step 4 Decision rule: Reject H0 if P-value<0.025 Step 5: Decision and conclusion: Since P-value<0.05, Reject H0 and conclude that and conclude that there is significant difference in decoding messages between the two method (Deciphermatic and QuickDecipher) Course #: MATH 2209 Course Name: Introduction to Statistics II Module One: The CLT, t-distributions and Inference for the Mean Assignment 1 Name: ID: 1. The weight of Skittles in a small bag is stated to be 61.5 g. The amount that the packaging machine puts in these bags is believed to have a Normal model with mean 65 g and standard deviation 1.8 g. a. What fraction of all bags sold are underweight? P(X<61.5)= P(Z<(61.5-65)/1.8) =P (Z<-1.9444)= 0.026 b. Some of the packages of Skittles are sold in \"bargain packs\" of three bags. What is the probability that none of the three is underweight? c. P(X>61.5) = P (Z<(61.5-65)/1.8/3) =P(Z>-3.3679) =0.9996 What is the probability that the mean weight of the three bags is below the stated amount? P(X<61.5) =1-0.9996=0.0004 d. What is the probability that the mean weight per bag of a 24-bag case of Skittles is below 65 g? P(X<65)=P(Z<65-65.5)/1.8/24)=P(X<0)=0.5 2. Receipts from a drug store show that customer purchases have a skewed distribution with a mean of $32 and a standard deviation of $20. a. Explain why you cannot determine the probability that the next customer will spend at least $40. Since the distribution is skewed, we cannot apply the probability that are gotten using properties of normal model. b. Can you estimate the probability that the next 10 customers will spend an average of at least $40? Explain. c. Is it likely that the next 50 customers will spend an average of at least $40? Explain. No, since the distribution is skewed, we need to a sample of at least 30 customers for the CLT to ensure a normal distribution for the sample means 3. A random sample of size ten was drawn from a population of 965 systolic blood pressure rates for males aged 15-20 years. The Minitab output showing the summary statistics for this sample of 10 is given below. Descriptive Statistics: C2 Variable C2 N 10 N* 0 Variable C2 Maximum 140.00 Mean 117.10 SE Mean 4.90 a. Identify the population and the sample. StDev 15.48 Minimum 90.00 Q1 104.00 Median 120.00 Q3 128.00 Course #: MATH 2209 Course Name: Introduction to Statistics II Module One: The CLT, t-distributions and Inference for the Mean b. Use the sample to calculate a 95% confidence interval for the average systolic blood pressure rate of all males. CI=x-bart*s/n CI=4.90 t*15.48/sqrt(10) (106.03, 128.17) c. Interpret this interval in the context of the problem. We are 95% confidence that the population mean is within (106.03, 128.17) d. At the 5% level of significance, does the data provide evidence that the mean systolic blood pressure rate is different from 120? i. State the hypotheses to be tested, and define the parameter. H0: =120 Ha: 120 ii. Check the required assumptions in the context of the question. Normality is met Sample size is less than 30 hence use t-distribution iii. Conduct the test and find the P-value P(X=120) T = (117.1- 120)/15.48/10) P-value=0.568 iv. State your conclusion in the context of the question. Reject H0 and conclude that the mean is different from 120 Course #: MATH 2209 Course Name: Introduction to Statistics II Module Two: Comparing Two Means Assignment 2 Name: ID: 1. In an effort to improve final grades in her class, a professor tried a new method of evaluation with one of her intro stats classes last term. A test of the null hypothesis that her old method (used with one class) and new method (used with her other class) don't differ in the mean final grade for students yielded a p-value of 0.066. Would a 95% confidence interval for new - old include 0? Explain. Yes, this is because p-value for the test is greater than 0.025 hence we expect the difference to be insignificant and thus the confidence interval will have a negative and a positive value hence including zero. 2. Students at a very large university have the option to take an introductory algebra class on campus or through distance learning. Dr. Brown teaches both the on campus and distance courses and gives all students the same final exam regardless of how they take the course. He is interested in whether there is a difference between the final exam grades depending on how the students take the course. He randomly selects 12 distance students and 12 on campus students. A Minitab output showing the data from his samples is below. Two-Sample T-Test and CI: Distance, On Campus Two-sample T for Distance vs On Campus Distance On Campus N 12 12 Mean 66.17 80.6 StDev 4.91 11.0 SE Mean 1.4 3.2 Difference = (Distance) - (On Campus) Estimate for difference: -14.42 95% CI for difference: (-21.85, -6.98) a. Interpret this interval in the context of the question. Suppose we repeated this study with different random samples for the two groups. Based on the confidence interval, we would expect 95% of the observed difference in sample means to be between (-21.85, -6.98) Also it can be interpreted as We are 95% confidence that the difference between the final exam grades depending on how the students take the course is within this interval b. What does this interval suggest about the difference between the two courses? Since 0 is not within the interval, thee is difference between the final exam grades depending on how the students take the course 3. Military intelligence units are often responsible for intercepting and deciphering coded messages used by enemy forces. In this type of work, time is critically important. Recently, the army compared two methods of deciphering coded messages. Two special agents randomly assigned 49 messages to two groups, and used identical computers to decode the messages. One of them used Deciphermatic (sample 1) to decode 24 messages and the other used QuickDecipher (sample 2) to decode 25 messages. The means and standard deviations for the time required by the computer, in minutes to decode the messages are given in the Minitab output below. Course #: MATH 2209 Course Name: Introduction to Statistics II Module Two: Comparing Two Means Test an appropriate hypothesis and use the p-value to state a conclusion. Be sure to follow the full steps to hypothesis testing, in the context of the question. Two-Sample T-Test and CI Sample N Mean StDev 1 24 37.48 8.16 2 25 31.02 3.68 SE Mean 1.7 0.74 Difference = mu (1) - mu (2) Estimate for difference: 6.46 95% CI for difference: (2.75, 10.17) T-Test of difference = 0 (vs not =): T-Value = 3.55 P-Value=0.001 DF=31 Answer Step 1: H0 1=2 Ha12 =0.05 Step 2: Test statistics=3.55 Step 3: P-value=0.001 Step 4 Decision rule: Reject H0 if P-value<0.025 Step 5: Decision and conclusion: Since P-value<0.05, Reject H0 and conclude that and conclude that there is significant difference in decoding messages between the two method (Deciphermatic and QuickDecipher)