Question

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.

Answer 1

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

`df=15+17-2=30`

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

`p_v =2*P(t_(30)>1.65) =0.109`

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

Explanation:

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

`\sigma^2_1 =\sigma^2_2 =\sigma^2`

And the statistic is given by this formula:

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

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

`\S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

This last one is an unbiased estimator of the common variance
`\sigma^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 = \mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 = 0`

Alternative hypothesis:
`\mu_1 -\mu_2 \\eq 0`

Our notation on this case :

`n_1 =15` represent the sample size for group 1

`n_2 =17` represent the sample size for group 2

`\bar X_1 =350` represent the sample mean for the group 1

`\bar X_2 =342` represent the sample mean for the group 2

`s_1=12` represent the sample standard deviation for group 1

`s_2=15` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`\S^2_p =((15-1)(12)^2 +(17 -1)(15)^2)/(15 +17 -2)=187.2`

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

`S_p=13.682`

And now we can calculate the statistic:

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

Now we can calculate the degrees of freedom given by:

`df=15+17-2=30`

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

`p_v =2*P(t_(30)>1.65) =0.109`

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