Question

Answer the following questions showing all work. Full credit will not be given to answers without work shown. If you use Minitab Express or StatKey include the appropriate output (copy + paste). If you do hand calculations show your work using the Word equation editor. Clearly identify your final answers. Output without explanation will not receive full credit and answers with no output or explanation will not receive full credit. Round all answers to 3 decimal places. If you have any questions, post them to the course discussion board.

1. A STAT 200 instructor wants to know if her students tend to score higher on the midterm exam than on the final exam. Data were collected from a representative sample of 46 students during the Fall 2017 semester. Data were paired by student. The mean difference, computed as midterm - final, was 1.3632 points with a standard deviation of 3.6362 points. [55 points]

A. In Minitab Express, conduct a paired means t test to determine if there is evidence that midterm exam scores are higher than final exam scores in the population of all STAT 200 students. Use the five-step hypothesis testing procedure and remember to include all relevant Minitab Express output. You should not need to do any hand calculations.

Step 1: Check assumptions and write hypotheses

Step 2: Calculate the test statistic

Step 3: Identify the p value

Step 4: Make a decision

Step 5: State a “real world” conclusion

B. In Minitab Express, conduct a single sample mean t test given a sample size of 46, sample mean of 1.3632, and sample standard deviation of 3.6362 to determine if there is evidence that the population mean is greater than 0. Use the five-step hypothesis testing procedure and remember to include all relevant Minitab Express output. You should not need to do any hand calculations.

Step 1: Check assumptions and write hypotheses

Step 2: Calculate the test statistic

Step 3: Identify the p value

Step 4: Make a decision

Step 5: State a “real world” conclusion

C. Explain why your test statistic and p-value were the same in parts B and C.

D. What minimum sample size would be necessary to construct a 95% confidence interval for the mean difference in midterm and final exam scores with a margin of error of 0.5 point? Show all work using the equation editor.

2. Download the BodyTemp.MTW file from Canvas. We will be comparing the body temperatures of men and women. [30 points]

A. Make a graph to compare the distributions of men and women’s body temperatures.

B. Use Minitab Express to determine if there is evidence that the mean body temperatures of men and women are different. The coding of gender is 0=man and 1=woman. Assume that the distribution of the body temperature data is normal. Use the five-step hypothesis testing procedure and remember to include all relevant Minitab Express output. You should not need to do any hand calculations.

Step 1: Check assumptions and write hypotheses

Step 2: Calculate the test statistic

Step 3: Identify the p value

Step 4: Make a decision

Step 5: State a “real world” conclusion

3. A random sample of Penn State World Campus undergraduate and graduate students were contacted and data concerning their gender identities were recorded. In a random sample of 40 undergraduate students, 21 identified as men and 19 identified as women. In a random sample of 40 graduate students, 17 identified as men and 23 identified as women. [15 points]

A. Is it appropriate to use the normal approximation method here to construct a confidence interval for the difference in population proportions? Show your work.

B. Construct a 95% confidence interval to compare the proportion of undergraduate students who identify as men to the proportion of graduate students who identify as men. If assumptions were met in part A, use the normal approximation method. Do not do any calculations by hand. Use Minitab Express and remember to copy+paste all relevant output and to clearly identify your final answer.

Answer 1

1)

A) The paired sample t test is done in minitab by following these steps,

Step 1: Stat > Basic Statistic > Paired t > OK. The screenshot is shown below,

Step 2: Select Summarized data (difference), enter Sample size: 46, Sample mean: 1.3632, Standard deviation: 3.6362 and click Options and select Alternative hypothesis: Difference > hypothesized difference then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

Now the five step model of hypothesis testing is describe below,

Step 1: The null hypothesis for the test is define as,

Null hypothesis: such that group means are all equal.

Step 2: The alternative hypothesis for the test is define as,

Alternative hypothesis, : such that means are not equal.

Step 3: The significance level for the test, is the maximum acceptance level at 5% risk which means null hypothesis will be rejected beyond this level ( such that if P-value is less than 0.05 null hypothesis will be rejected)

Step 4: The test statistic is the T-value which is obtained in Paires sample T-Test summary, The T-value is used to calculate the P-value. Here the P-value is used to make a decision about the statistical significance of the difference in means. From the result summary,

Step 5: The decision for the statistical significance in mean is determine by the P-value. The P-value can be define as the probability that measures the evidence against the null hypothesis If the P-value < alpha = 0.05: The differences between some of the means are statistically significant and if P-value > alpha = 0.05: The differences between the means are not statistically significant. from the model summary,

Hence the difference between the means is statistically significant.

B)

The one sample t test is done in minitab by following these steps,

Step 1: Stat > Basic Statistic > 1 Sample t > OK. The screenshot is shown below,

Step 2: Select Summarized data (difference), enter Sample size: 46, Sample mean: 1.3632, Standard deviation: 3.6362 and tick Perform hypothesis test click Options and select Alternative hypothesis: Mean > hypothesized mean then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

Now the five step model of hypothesis testing is describe below,

Step 1: The null hypothesis for the test is define as,

Null hypothesis: such that group means are all equal.

Step 2: The alternative hypothesis for the test is define as,

Alternative hypothesis, : such that means are not equal.

Step 3: The significance level for the test,

Step 4: The test statistic is the T-value which is obtained in one sample T-Test summary, The T-value is used to calculate the P-value. Here the P-value is used to make a decision about the statistical significance of the difference in means. From the result summary,

Step 5: The decision for the statistical significance in mean is determine by the P-value. From the model summary,

Hence the means are statistically significant.

C) The test statistic for paired t test and one sample t test is same beacuse the sample mean and the standard deviations are same.

D) The sample size is obtained using the formula,

2)

Data file is required to solve the problem

3)

A) If the sample size n > 30, we can take the normal approximation assumption. Here the sample size n = 40. hence the assumption is valid.

B) The confidence interval for proportion of men in undergraduate is obtained in minitab by following these steps,

Step 1: Stat > Basic Statistic > One-Sample Proportion > OK. The screenshot is shown below,

Step 2: Select Summarized data, Number of events: 21, Number of trials: 40 then and click Option: Select Confidence level: 95.0, Method: Normal approximation then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below

Similarly the confidence interval for graduate students is obtained as follow,

Step 1: Stat > Basic Statistic > One-Sample Proportion > OK. The screenshot is shown below,

Step 2: Select Summarized data, Number of events: 17, Number of trials: 40 then and click Option: Select Confidence level: 95.0, Method: Normal approximation then OK. The screenshot is shown below,

The result is obtained, The screenhot is shown below,

For under graduate

95% CI for p

(0.370245, 0.679755)

For graduate

95% CI for p

(0.271804, 0.578196)