### Analysis of Proctored vs. Nonproctored TestsA study was conducted on proctored and nonproctored tests. The results are summarized in the following table. The assumption is that two independent simple random samples are taken from normally distributed populations, without assuming equal population standard deviations. Complete parts (a) and (b) below.#### Data Summary:| | Proctored (\( \mu_1 \)) | Nonproctored (\( \mu_2 \)) ||--------------------|-------------------------|----------------------------|| Sample Size (\( n \)) | 32 | 33 || Mean (\( \bar{x} \)) | 78.49 | 84.99 || Standard Deviation (\( s \)) | 11.61 | 18.83 |#### Hypothesis TestingChoose the appropriate 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.** \( H_0: \mu_1 = \mu_2 \) \\ \( H_1: \mu_1 \neq \mu_2 \) **Selection:** **C.**#### Test Statistic:- The test statistic, \( t \), is \( -0.43 \). (Rounded to two decimal places as needed.)#### P-value:- The P-value is \( 0.337 \). (Rounded to three decimal places as needed.)#### Conclusion of the Test:State the conclusion:- **A.** Reject \( H_0 \). There is sufficient evidence to support that students taking nonproctored tests achieve higher mean scores than those taking proctored tests.- **B.** Reject \( H_0 \). There is not sufficient evidence to support that students taking nonproctored tests achieve higher mean scores than those taking proctored tests

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Description: ### Analysis of Proctored vs. Nonproctored TestsA study was conducted on proctored and nonproctored tests. The results are summarized in the following table. The assumption is that two independent simple random samples are taken from normally distri

(b) The degrees of freedom is, Df=smaller of(n1-1)(n2-1) =32-1 =31 The critical-value is obtained by using Excel function “=TINV(probability, degrees of freedom)”. Output obtained from Excel is given below: From the output, the t* is 2.039.The 95% confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests is (-14.39, 1.39), which is obtained below: CI=x¯1-x¯2±t*s12n1+s22n2=78.49-84.99±2.03911.61232+18.83333=-6.5±7.8856=-14.3856,1.3856=-14.39,1.39 Here, the hypothesized value 0 contains in the 95% confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Hence, it can be concluded that there is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Yes, the confidence interval support the conclusion of the test. Yes, because the confidence interval contains the hypothesized value 0 .

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Proctored Nonproctored 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. H2 33 84.99 18.83 32 x 78.49 11.61 O B. Ho: H1 = H2 O A. Ho: H1 # H2 H1: H1 H2 O C. Ho: H1 = H2 H;: H1

Proctored Nonproctored 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. H2 33 84.99 18.83 32 x 78.49 11.61 O B. Ho: H1 = H2 O A. Ho: H1 # H2 H1: H1 H2 O C. Ho: H1 = H2 H;: 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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