Answer to Question #235411 in Statistics and Probability for nat

Question

 Consider the following hypotheses and the sample data drawn independently from two normally distributed populations: H0 : 1 2 D 0 against H1 : 1 2 > 0 A random sample of n1 D 20 is selected with a mean of 57 and a second random sample of n2 D 20 is independently selected with a mean of 63. Assume that the two samples are normally distributed and the population standard deviations are 11:5 and 15:2 respectively. The analyst wants to test whether the population means differ at 5% significance level. The p-value is 1. 0:0793 2. 0:0808 3. 0:9207 4. 1:4078 5. 0:

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1-\mu_2=0$

$H_0:\mu_1-\mu_2>0$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=df_1+df_2 =20-1+20-1=38$$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{57-63}{\sqrt{\dfrac{(20-1)(11.5)^2+(20-1)(15.2)^2}{20+20-2}(\dfrac{1}{20}+\dfrac{1}{20})}}$$

$$\approx-1.4078$$

The p-value for right-tailed $\alpha=0.05, df=38, t=-1.4078$ is $p=0.91634.$

The degrees of freedom are computed as follows, assuming that the population variances are unequal:

$$df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}$$

$$=\dfrac{(\dfrac{(11.5)^2}{20}+\dfrac{(15.2)^2}{20})^2}{\dfrac{((11.5)^2/20)^2}{20-1}+\dfrac{(s_2^2/20)^2}{20-1}}$$

$$\approx35.3835$$

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

$$=\dfrac{57-63}{\sqrt{\dfrac{(11.5)^2}{20}+\dfrac{(15.2)^2}{20}}}$$

$$\approx-1.4078$$

The p-value for right-tailed $\alpha=0.05, df=35.3835, t=-1.4078$ is $p=0.916041.$