2. research was being done on who makes more money right after college, a person with a sociology degree or…

2. research was being done on who makes more money right after college, a person with a sociology degree or a psychology degree. to test this, the researcher randomly selected 45 recent sociology graduates and found that the mean income they are receiving is $53,900 with a standard deviation of $2,050. the researcher also randomly selected 40 psychology majors and found that the mean income they are receiving is $55,200 with a standard deviation of $3,850. test the claim that the mean income for psychology majors is more than the mean income for sociology majors at the $alpha = 0.05$ level of significance.

2. research was being done on who makes more money right after college, a person with a sociology degree or a psychology degree. to test this, the researcher randomly selected 45 recent sociology graduates and found that the mean income they are receiving is $53,900 with a standard deviation of $2,050. the researcher also randomly selected 40 psychology majors and found that the mean income they are receiving is $55,200 with a standard deviation of $3,850. test the claim that the mean income for psychology majors is more than the mean income for sociology majors at the $alpha = 0.05$ level of significance.

Answer

Explanation:

Step1: State the hypotheses

Let $\mu_1$ be the mean income of sociology majors and $\mu_2$ be the mean income of psychology majors. $H_0:\mu_1\geq\mu_2$ (null hypothesis), $H_1:\mu_1 < \mu_2$ (alternative hypothesis)

Step2: Calculate the test - statistic

The formula for the two - sample independent t - test statistic when the population variances are unknown and assumed unequal is: [t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}] where $\bar{x}_1 = 53900$, $s_1 = 2050$, $n_1 = 45$, $\bar{x}_2=55200$, $s_2 = 3850$, $n_2 = 40$, and $\mu_1-\mu_2 = 0$ under $H_0$. [t=\frac{(53900 - 55200)-0}{\sqrt{\frac{2050^{2}}{45}+\frac{3850^{2}}{40}}}] [=\frac{- 1300}{\sqrt{\frac{4202500}{45}+\frac{14822500}{40}}}] [=\frac{-1300}{\sqrt{93388.89 + 370562.5}}] [=\frac{-1300}{\sqrt{463951.39}}] [=\frac{-1300}{681.14}] [t\approx - 1.91]

Step3: Determine the degrees of freedom

The formula for the degrees of freedom for the two - sample independent t - test with unequal variances is: [df=\frac{\left(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}\right)^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}] [=\frac{\left(\frac{2050^{2}}{45}+\frac{3850^{2}}{40}\right)^{2}}{\frac{(2050^{2}/45)^{2}}{44}+\frac{(3850^{2}/40)^{2}}{39}}] [=\frac{(93388.89+370562.5)^{2}}{\frac{(93388.89)^{2}}{44}+\frac{(370562.5)^{2}}{39}}] [df\approx 57]

Step4: Find the critical value

For a one - tailed test with $\alpha = 0.05$ and $df = 57$, using a t - distribution table or a calculator, the critical value $t_{\alpha,df}=t_{0.05,57}\approx1.672$

Step5: Make a decision

Since our calculated $t=-1.91$ and the critical value is $1.672$, and we have a one - tailed test with the alternative hypothesis $H_1:\mu_1 < \mu_2$, we reject the null hypothesis $H_0$ because $- 1.91<1.672$

Answer:

We reject the null hypothesis. There is sufficient evidence at the $\alpha = 0.05$ level of significance to claim that the mean income for psychology majors is more than the mean income for sociology majors.