Parts a and b please (a) Test whether u, > H, at the a = 0.01 level of significance for the given sample data.(b) Construct a 99% confidence interval about u,-H2.

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Description: Parts a and b please (a) Test whether u, > H, at the a = 0.01 level of significance for the given sample data.(b) Construct a 99% confidence interval about u,-H2.

a) The provided sample means are shown below: Xˉ1=48.2\bar X_1 = 48.2 Xˉ2=46.1\bar X_2 = 46.1…

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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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Parts a and b please

(a) Test whether u, > H, at the a = 0.01 level of significance for the given sample data.
(b) Construct a 99% confidence interval about u,-H2.

Expert Solution

Step 1

The provided sample means are shown below:

$\bar X_1 = 48.2$

$\bar X_2 = 46.1$

Also, the provided sample standard deviations are:

$s_1 = 5.5$

$s_2 = 11.8$

and the sample sizes are $n_1 = 20$ and $n_2 = 17$

Step 2

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: $\mu_1$ = $\mu_2$

Ha: $\mu_1$ > $\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, will be used.

Rejection Region

Based on the information provided, the significance level is $\alpha = 0.01$ , and the degrees of freedom are $df = 35$ . In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this right-tailed test is $t_c = 2.438$ for $\alpha = 0.01$ and $df = 35$

The rejection region for this right-tailed test is $R = {t : t > 2.438}$ .

Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$= \frac{ 48.2 - 46.1}{\sqrt{ \frac{(20-1)5.5^2 + (17-1)11.8^2}{ 20+17-2}(\frac{1}{ 20}+\frac{1}{ 17}) } } = 0.711$

Decision about the null hypothesis

Since it is observed that $t = 0.711 \le t_c = 2.438$ , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p = 0.2408$ , and since $p = 0.2408 \ge 0.01$ , it is concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2$ , at the 0.01 significance level.

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