Answer to Question #214662 in Statistics and Probability for danigurl

Question

A study is conducted to compare the performance of students with more than one personal electronic gadget and those with only one. A few of them were taken as respondents for the study. The mean grades of these students and the standard deviations are shown below. Is it possible to conclude that there is no significant difference in the mean grades of the two types of students at a 95%

confidence level?

Students Mean Standard Deviation Sample Size

With one gadget xÌ…1 = 83 s1 = 12 n1 = 7

With more than one gadget xÌ…2 = 79 s2 = 13 n2 = 5

Answer 1

First we test the equality of variances.

$H_0:\sigma_1^2=\sigma_2^2$

$H_a:\sigma_1^2\ne\sigma_2^2$

$F=\frac{s_1^2}{s_2^2}$

$=\frac{144}{169}$

$=0.852$

$cv=F_{0.05,6,4}=6.16$

$6.16>0.852$, thus, we reject the null hypothesis. Variances are significantly different.

The samples are independent with unequal variances($S_1^2=144,s_2^2=169)$ ; thus, two independent samples t-test assuming unequal variances is appropriate. Assuming unequal variances works whether variances are equal or not.

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\ne\mu_2$

$t = \frac{\bar{X_{1}} - \bar{X_{2}}} {\sqrt{\frac{{s^{2}_{1}}}{n_{1}} + \frac{{s^{2}_{2}}}{n_{2}}}}$

$df = \frac{(\frac{s^{2}_{1}}{n_{1}} + \frac{s^{2}_{2}}{n_{2}})^{2}} {\frac{(\frac{s^{2}_{1}}{n_{1}})^{2}}{n_{1}-1 }+ \frac{(\frac{s^{2}_{2}}{n_{2}})^{2}}{n_{2}-1 }}$

$t = \frac{{83} - 79} {\sqrt{\frac{{144}}{7} + \frac{169}{5}}}$

$=0.5425$

$df = \frac{(\frac{144}{7} + \frac{169}{5})^{2}} {\frac{(\frac{144}{7})^2}{6 }+ \frac{(\frac{169}{5})^2}{4 }}$

$=8.3\approx8$

$cv=t_{0.025,8}=\pm2.306$

Since the test statistic (0.5425) is less than the critical value (2.306), we fail to reject the null hypothesis. There is no significant difference in the mean grades of students with more than one personal electronic gadget and those with only one.