Question

Two random samples of 40 students were drawn independently from two populations of students. Assume their aptitude tests are normally distributed (total points = 100). The following statistics regarding their scores in an aptitude test were obtained:xbar1 = 76 s1= 8 xbar2 = 72 and s2= 6.5. Do not assume population variances are equal. Use all formulas assuming unequal variances.
Test at the 5% significance level to determine whether we can infer that the two population means differ.

Explain how to use the 95% confidence interval to test the hypotheses at a = .05.

Answer 1

Answer:

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the population mean for group 1 is significantly different than the population mean for group 2.  

The 95% confidence interval is given by (0.756;7.243), since the interval not contain's the 0 we can reject the null hypothesis that the means are equal.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=76$ represent the mean for 1  

$\bar X_{2}=72$ represent the mean for 2  

$s_{1}=8$ represent the sample standard deviation for 1  

$s_{2}=6.5$ represent the sample standard deviation for 2

$n_{1}=40$ sample size for the group 1  

$n_{2}=40$ sample size for the group 2  

$\alpha=0.05$ Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the population means differs, the system of  hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}= 0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

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

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

$t=\frac{(76-72)-0}{\sqrt{\frac{8^2}{40}+\frac{6.5^2}{40}}}}=2.454$  

P value  

We need to find first the degrees of freedom given by:

$df=n_1 +n_{2}-2=40+40-2=78$

Since is a two tailed test the p value would be:  

$p_v =2*P(t_{78}>2.454)=0.016$  

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the population mean for group 1 is significantly different than the population mean for group 2.  

Confidence interval

The critical value for this case can be calculated like this. $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. The degrees of freedom are 78 so we can use this code in excel to find the critical value for the interval "=-T.INV(0.025,78)" and we got $t_{\alpha/2}=1.99$

The confidence interval for this case would be given by this formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

And if we replace the values given we have:

$(76 -72) - 1.99\sqrt{\frac{8^2}{40}+\frac{6.5^2}{40}}}=0.756$

$(76 -72) +1.99\sqrt{\frac{8^2}{40}+\frac{6.5^2}{40}}}=7.243$

The 95% confidence interval is given by (0.756;7.243), since the interval not contain's the 0 we can reject the null hypothesis that the means are equal.