Chapter 10.1, Problem 10.28E

(a)

To determine

To Explain: these data satisfy our guidelines for safe use of the chi-square test.

Answer to Problem 10.28E

Yes

Explanation of Solution

Given:

	Female	Male
Accounting	68	56
Administration	91	40
Economics	5	6
Finance	61	59

Formula used:

  $E=np$

Calculation:

Assume

  $\alpha =0.05=5\%$

Row and column totals are

	Female	Male	Total
Accounting	68	56	124
Administration	91	40	131
Economics	5	6	11
Finance	61	59	120
Total	225	161	386

The expected frequencies $E$

  $\begin{array}{c}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{124\times 225}{386}=72.2798\\ E_{12}=\frac{r_1\times c_2}{n}=\frac{124\times 168}{386}=15.7202\\ E_{21}=\frac{r_2\times c_1}{n}=\frac{131\times 225}{386}=76.3601\\ E_{22}=\frac{r_2\times c_2}{n}=\frac{131\times 161}{386}=54.6399\\ E_{31}=\frac{r_3\times c_1}{n}=\frac{11\times 161}{386}=6.4119\\ E_{32}=\frac{r_3\times c_2}{n}=\frac{11\times 161}{386}=4.5881\\ E_{41}=\frac{r_4\times c_1}{n}=\frac{120\times 225}{386}=69.9482\\ E_{42}=\frac{r_4\times c_2}{n}=\frac{120\times 161}{386}=50.0518\end{array}$

It is observed that all expected counts are least 1 and less than 20% of the expected counts are less than 5 therefore the data satisfied our guidelines for safe use of the chi-square test.

(b)

To determine

To Explain: there is statistically significant relationship between the gender and major of business students and carry out a test and the conclusion.

Answer to Problem 10.28E

There is no enough evidence to help the claim that there is an connection between major and gender.

Explanation of Solution

Given:

	Female	Male
Accounting	68	56
Administration	91	40
Economics	5	6
Finance	61	59

Formula used:

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

The null hypothesis statement is that there is no connection between the variables, where the alternative hypothesis statement is that there is a connection between the variable.

  $H_0:$ There is no connection between Gender and Major.

  $H_1:$ There is a connection between Gender and Major.

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(68-72.2798\right)}^2}{72.2798}+\frac{{\left(56-51.7202\right)}^2}{51.7202}\\ +\frac{{\left(91-76.3601\right)}^2}{76.3601}+\frac{{\left(40-54.6399\right)}^2}{54.6399}\\ +\frac{{\left(5-6.4119\right)}^2}{6.4119}+\frac{{\left(6-4.5881\right)}^2}{4.5881}\\ +\frac{{\left(61-69.9482\right)}^2}{69.9482}+\frac{{\left(59-50.0518\right)}^2}{50.0518}\\ =10.8267\end{array}$

The degree of freedom is

  $df=(r-1)\left(c-1\right)=\left(4-1\right)\left(2-1\right)=3$

The P-Value is the probability of obtaining the value of the test statistic, or a value more extreme and row $df=3:$

  $0.01<P<0.05$

If the P-value is equal or less than to the significance level, then the alternative hypothesis is accepted and null hypothesis is rejected:

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is no enough evidence to help the claim that there is an connection between major and gender.

(c)

To determine

To Calculate: the percent of the students did not respond to the questionnaire and conclusion that can be drawn from these data.

Answer to Problem 10.28E

46.54% the results could be strongly inaccurate

Explanation of Solution

The table have total data about 386 students, where 722 questionnaires were sending out. Then observed that 722-386=336 of the students did not respond to the questionnaires

  $\frac{336}{722}=0.4654=46.54\%$

The results could be strongly inaccurate.