How do you solve for B on a ti-84 plus calc **Statistical Analysis of Proctored vs. Nonproctored Test Scores**A study was conducted to compare the scores of students taking proctored versus nonproctored tests. The dependent variable is test scores, and the independent variable is the type of test administration (proctored or nonproctored). The results are summarized in the table below. Note that the population standard deviations are not assumed to be equal.**Table: Summary of Test Results**| | Proctored (Group 1) | Nonproctored (Group 2) ||------------------|---------------------|------------------------|| Number of Samples (n) | 35 | 30 || Mean Score (x̄) | 77.85 | 82.25 || Standard Deviation (s) | 10.06 | 22.41 |**a. Hypothesis Testing at \(\alpha = 0.05\)**To test the claim that students taking nonproctored tests achieve a higher mean score than those taking proctored tests, we formulate the hypotheses as follows:- Null Hypothesis (\(H_0\)): \(\mu_1 = \mu_2\)- Alternative Hypothesis (\(H_1\)): \(\mu_1 < \mu_2\)**Test Statistic Calculation**The test statistic \( t_c \) is:\[ t_c = -0.99 \]**P-Value Calculation**The P-value is:\[ P = 0.163 \]**Conclusion**To conclude the hypothesis test, we compare the P-value to the significance level \(\alpha\):- Because \(P > \alpha\), we fail to reject the null hypothesis \(H_0\).**Conclusion:** There is not sufficient evidence to support the claim that students taking nonproctored tests achieve a higher mean score than those taking proctored tests.**b. Confidence Interval for the Difference in Means**We construct a confidence interval for the difference in means between the two groups:\[ -12.81 < \mu_1 - \mu_2 < 4.01 \]This interval has been rounded to two decimal places.**Interpretation:** Since the 95% confidence interval includes zero, this further substantiates the conclusion that there is no significant difference in the mean scores between students taking proctored and non

(B) (b) Obtain the 95% confidence interval for the difference between two means. The 95%…

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. a. Use a 0.05 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? Ạ Họ bây g Hy: Hy #4₂ C. Ho: H₁ H2 H₁: Hy

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. a. Use a 0.05 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? Ạ Họ bây g Hy: Hy #4₂ C. Ho: H₁ H2 H₁: Hy

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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