Question

Given the following null and alternative hypotheses H0: μ1 ≥ μ2 HA: μ1 < μ2 Together with the following sample information (shown below). Assuming that the populations are normally distributed with equal variances, test at the 0.10 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach. Sample 1 Sample 2 n1 = 14 n2 = 18 x-bar1 = 565 x-bar2 = 578 s1 = 28.9 s2 = 26.3

Answer 1

Null hypothesis:
`\mu_1 \geq \mu_2`

Alternative hypothesis:
`\mu_1 <\mu_2`

`t=-1.329`

`p_v =P(t_(30)<-1.329) =0.0969`

With the p value obtained and using the significance level given
`\alpha=0.1` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.

Explanation:

When we have two independnet 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
`\simga^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 \geq \mu_2`

Alternative hypothesis:
`\mu_1 <\mu_2`

Or equivalently:

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

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

Our notation on this case :

`n_1 =14` represent the sample size for group 1

`n_2 =18` represent the sample size for group 2

`\bar X_1 =565` represent the sample mean for the group 1

`\bar X_2 =578` represent the sample mean for the group 2

`s_1=28.9` represent the sample standard deviation for group 1

`s_2=26.3` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`\S^2_p =((14-1)(28.9)^2 +(18 -1)(26.3)^2)/(14 +18 -2)=753.882`

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

`S_p=27.457`

And now we can calculate the statistic:

`t=\frac{(565 -578)-(0)}{27.457\sqrt{(1)/(14)}+(1)/(18)}=-1.329`

Now we can calculate the degrees of freedom given by:

`df=14+18-2=30`

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

`p_v =P(t_(30)<-1.329) =0.0969`

So with the p value obtained and using the significance level given
`\alpha=0.1` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.