Question

1.   A survey is conducted on attitudes towards drinking. A random sample of eight married couples is selected, and the husbands and wives respond to an attitude-toward-drinking scale. (20 Points)The scores are as follows:

HUSBAND SCORE - 16 , 20, 10, 15, 8, 19, 14, 15

WIFE SCORE - 15, 18, 13, 10, 12, 16, 11, 12

Answer 1

$$\bar{x}_1=\dfrac{1}{8}(16+ 20+10+15+8+19+14+15)$$

$$=14.625$$

$$s_1^2=\dfrac{1}{8-1}((16-14.625)^2+ (20-14.625)^2$$

$$+(10-14.625)^2+(15-14.625)^2+(8-14.625)^2$$

$$+(19-14.625)^2+(14-14.625)^2+(15-14.625)^2)$$

$$=\dfrac{115.875}{7}\approx16.5535$$

$$s_1=\sqrt{s_1^2}\approx4.0686$$

$$\bar{x}_2=\dfrac{1}{8}(15+ 18+13+10+12+16+11+12)$$

$$=13.375$$

$$s_2^2=\dfrac{1}{8-1}((15-13.375)^2+ (18-13.375)^2$$

$$+(13-13.375)^2+(10-13.375)^2+(12-13.375)^2$$

$$+(16-13.375)^2+(11-14.625)13.375^2+(12-13.375)^2)$$

$$=\dfrac{51.875}{7}\approx7.4107$$

$$s_2=\sqrt{s_2^2}\approx2.7223$$

A F-test is used to test for the equality of variances.

The critical values for a significance level of $\alpha=0.05,$ and degrees of freedom $df_1=7$

and $df_2=7$ are $F_U=F_{1-\alpha/2; df_1, df_2}=4.995$ and $F_L=F_{\alpha/2; df_1, df_2}=0.2.$

The F-statistic is calculated by dividing the sample variances, as shown below:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{4.0686}{2.7233}\approx2.2337$$

Since $F_L=0.2<F=2.2337<F_U=4.995,$ then we fail to reject the null hypothesis of equal population variances, and hence, we assume that the population variances are equal. 

1. We need to construct the 95% confidence interval for the difference between the population means $\mu_1-\mu_2,$ for the case that the population standard deviations are not known. 

We assume that the population variances are equal. Therefore, the corresponding degrees of freedom are computed as follows:

$$df=df_1+df_2=7+7=14$$

The critical value for $\alpha=0.05$ and $df=14$ is $t_c=2.144787.$

The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

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

$$=\sqrt{\dfrac{(8-1)\dfrac{115.875}{7}+(8-1)\dfrac{51.875}{7}}{8+8-2}}\approx3.4615$$

Since we assume that the population variances are equal, the standard error is computed as follows:

$$SE=s_p\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}=3.4615\sqrt{\dfrac{1}{8}+\dfrac{1}{8}}\approx1.7308$$

Now, we finally compute the confidence interval:

$$CI=(\bar{x_1}-\bar{x_2}-t_c\times SE, \bar{x_1}-\bar{x_2}+t_c\times SE)$$

$$=(14.625-13.375-2.144787\times 1.7308,$$

$$14.625-13.375+2.144787\times 1.7308)$$

$$=(-2.4622,4.9622)$$

Therefore, based on the data provided, the 95% confidence interval for the difference between the population means $\mu_1-\mu_2$ is $-2.4622<\mu_1-\mu_2<4.9622,$ which indicates that we are 95% confident that the true difference between population means is contained by the interval $(-2.4622,4.9622).$

2. The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

The rejection region for this two-tailed test is $R=\{t:|t|>2.144787\}.$

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

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}=0.722$$

Since it is observed that $|t|=0.722<2.144787=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.05, df=14, t=0.722$ is $p=0.482181,$ and since $p=0.482181>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha=0.05$ significance level.