A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal.**Table:**- **Proctored** - \( \mu_1 \) - \( n = 32 \) - \( \bar{x} = 78.49 \) - \( s = 11.61 \)- **Nonproctored** - \( \mu_2 \) - \( n = 33 \) - \( \bar{x} = 84.99 \) - \( s = 18.83 \)---**a.** Use a 0.01 significance level to test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.**What are the null and alternative hypotheses?**- **A.** \( H_0: \mu_1 \neq \mu_2 \) \( H_1: \mu_1 < \mu_2 \)- **B.** \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 > \mu_2 \)- **C.** \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 < \mu_2 \)- **D.** \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 \neq \mu_2 \)**The test statistic, t, is \_\_\_. (Round to two decimal places as needed.)****The P-value is \_\_\_. (Round to three decimal places as needed.)****State the conclusion for the test.**- **A.** Reject \( H_0 \). There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.- **B.** Reject \( H_0 \). There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean **Constructing a Confidence Interval for the Mean Score Difference**In this exercise, we aim to construct a confidence interval suitable for testing the claim that students taking nonproctored tests achieve a higher mean score than those taking proctored tests.### Confidence Interval ConstructionWe are asked to find the range for the difference in means:\[ \_\_\_ < \mu_1 - \mu_2 < \_\_\_ \]*Note: Please round your answers to two decimal places as necessary.*### InterpretationOnce the confidence interval is calculated, determine if it supports the conclusion of the test:Fill in the blanks accordingly:- **Does the confidence interval support the conclusion of the test?** - [Dropdown] because the confidence interval contains [Dropdown] Complete the calculation and interpretation based on the provided statistical data.

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Description: A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standar

As per honor code, we will only answer first three subparts of the question. a) Perform the two sample t-test using the sample data. Null hypothesis: μ1=μ2 Alternative hypothesis: μ1<μ2 This hypotheses constitute a two-tailed test and significance level for the test is 0.01. Use the sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t). S.E=s12n1+s22n2=(11.61)232+(18.83)233=3.8674 D.F=n1+n2-2=32+33=63 t=(x1-x2)- dS.E=(78.49-84.99)-03.8674=-1.6807≈-1.68The P-value for this test is the probability that a t statistic having 63 degrees of freedom is less than -1.68. P-value = P(t < -1.68) Use the excel formula, "=T.DIST(-1.68,63,TRUE)" to determine the P-value. P-value = 0.049 Decision: Since the P-value (0.049) is greater than the significance level (0.01), hence we failed to reject the null hypothesis. Conclusion: (D) Fail to reject H0. There is not sufficient evidence to support the claim that students taking non proctored tests get a higher mean score than those taking proctored tests.

Math

Statistics

Proctored Nonproctore A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. H1 H2 33 32 X 78.49 84.99 Is 11.61 18.83 a. Use a 0.01 significance level to test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. What are the null and alternative hypotheses? O A. Ho: H1 # H2 O B. Ho: H1 = H2 H1: H1> H2 O D. Ho: H1 =H2 H1: H1 # H2 O C. Ho: H1 = H2 H1: H1

Proctored Nonproctore A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. H1 H2 33 32 X 78.49 84.99 Is 11.61 18.83 a. Use a 0.01 significance level to test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. What are the null and alternative hypotheses? O A. Ho: H1 # H2 O B. Ho: H1 = H2 H1: H1> H2 O D. Ho: H1 =H2 H1: H1 # H2 O C. Ho: H1 = H2 H1: H1

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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