A study was conducted to determine if the percent of women who receivedfinancial aid is different form the percentage of men who receive financial aid. Arandom sample of students revealed the results shown in the table. Using(a=0.01), is there significant evidence to reject the null hypothesis?womenmensample sizenumber receiving aid 200250.300225a. Find the 90% confidence interval for the difference in financial aid ratesbetween men and women and interpret this interval in the context of theproblem.b. Write the two hypotheses for the problem.Compute the appropriate statistical hypothesis test and report the teststatistic and p-value.d. Write the conclusion for the problem regarding the null hypothesis usingstatistical language.С.

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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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A study was conducted to determine if the percent of women who received
financial aid is different form the percentage of men who receive financial aid. A
random sample of students revealed the results shown in the table. Using
(a=0.01), is there significant evidence to reject the null hypothesis?
women
men
sample size
number receiving aid 200
250.
300
225
a. Find the 90% confidence interval for the difference in financial aid rates
between men and women and interpret this interval in the context of the
problem.
b. Write the two hypotheses for the problem.
Compute the appropriate statistical hypothesis test and report the test
statistic and p-value.
d. Write the conclusion for the problem regarding the null hypothesis using
statistical language.
С.

Expert Solution

Step 1

Given that,

Sample size of women $n_{1} = 250$

Sample size of men $n_{1} = 300$

Sample proportion of women receiving aid $\hat{p_{1}} = \frac{200}{250} = 0.80$

Sample proportion of men receiving aid $\hat{p_{2}} = \frac{225}{300} = 0.75$

Step 2

a. Confidence interval:

Hypothesis:

$H_{0} : p_{1} - p_{2} = 0$

$H_{1} : p_{1} - p_{2} \neq 0$

Critical value:

The two tailed z critical value at 90% confidence level is 1.645.

Calculation:

The 90% confidence interval for the difference in financial aid rates between men and women can be calculated as follows:

$\begin{array}{rcl} C I & = & (\hat{p_{1}} - \hat{p_{2}}) \pm z_{c} (\sqrt{\frac{\hat{p_{1}} (1 - \hat{p_{1}})}{n_{1}} + \frac{\hat{p_{2}} (1 - \hat{p_{2}})}{n_{1}}}) \\ = & (0.80 - 0.75) \pm 1.645 (\sqrt{\frac{0.80 (1 - 0.80)}{250} + \frac{0.75 (1 - 0.75)}{300}}) \\ = & 0.05 \pm 0.05851 \\ = & (- 0.0085, 0.1085) \end{array}$

Interpretation:

Since the value of the parameter specified by the null hypothesis (0) is contained in the 90% confidence interval, the null hypothesis cannot be rejected.

Thus, there is no sufficient evidence to conclude that the percent of women who received financial aids is different from the percent of men who received financial aids.

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