A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level,what is the p-value for comparing the means, given that the variances are equal.

If we compare the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]

Our notation on this case :

[tex]n_1 =15[/tex] represent the sample size for group 1

[tex]n_2 =17[/tex] represent the sample size for group 2

[tex]\bar X_1 =350[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =342[/tex] represent the sample mean for the group 2

[tex]s_1=12[/tex] represent the sample standard deviation for group 1

[tex]s_2=15[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]\S^2_p =\frac{(15-1)(12)^2 +(17 -1)(15)^2}{15 +17 -2}=187.2[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=13.682[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(350 -342)-(0)}{13.682\sqrt{\frac{1}{15}+\frac{1}{17}}}=1.65[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=15+17-2=30[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{30}>1.65) =0.109[/tex]

If we compare the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.