Question

construct a 95% confidence interval for $\mu_1 - \mu_2$. two samples are rando sample statistics are given below. assume that $\sigma_1^2 = \sigma_2^2$. $n_1 = 11, n_2 = 18, \bar{x}_1 = 4.8, \bar{x}_2 = 5.2, s_1 = 0.76, s_2 = 0.51$ a. $(-1.762, 1.762)$ b. $(-0.883, 0.083)$ c. $(-4.152, 3.981)$ d. $(-2.762, 2.762)$

Explanation:

Step1: Calculate pooled variance

The formula for pooled variance \( s_p^2 \) when \( \sigma_1^2=\sigma_2^2 \) is \( s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2} \).
Substitute \( n_1 = 11,s_1 = 0.76,n_2 = 18,s_2 = 0.51 \):
\( (n_1 - 1)s_1^2=(11 - 1)\times0.76^2=10\times0.5776 = 5.776 \)
\( (n_2 - 1)s_2^2=(18 - 1)\times0.51^2=17\times0.2601 = 4.4217 \)
\( n_1 + n_2-2=11 + 18-2 = 27 \)
\( s_p^2=\frac{5.776 + 4.4217}{27}=\frac{10.1977}{27}\approx0.3777 \)

Step2: Calculate standard error

The standard error \( SE \) for the difference in means is \( SE = s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \), where \( s_p=\sqrt{s_p^2}=\sqrt{0.3777}\approx0.6146 \)
\( \frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{11}+\frac{1}{18}=\frac{18 + 11}{198}=\frac{29}{198}\approx0.1465 \)
\( SE = 0.6146\times\sqrt{0.1465}\approx0.6146\times0.3828\approx0.2353 \)

Step3: Calculate degrees of freedom and t - critical value

Degrees of freedom \( df=n_1 + n_2-2 = 27 \). For a 95% confidence interval, \( \alpha = 0.05 \), two - tailed, the critical value \( t_{\alpha/2,df}=t_{0.025,27}\approx2.052 \) (from t - distribution table)

Step4: Calculate the difference in sample means and margin of error

The difference in sample means \( \bar{x}_1-\bar{x}_2=4.8 - 5.2=- 0.4 \)
Margin of error \( E=t_{\alpha/2,df}\times SE=2.052\times0.2353\approx0.483 \)

Step5: Calculate confidence interval

The confidence interval for \( \mu_1-\mu_2 \) is \( (\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E \)
Substitute values: \( - 0.4-0.483<\mu_1 - \mu_2<-0.4 + 0.483 \)
\( - 0.883<\mu_1 - \mu_2<0.083 \)

Answer:

B. \((-0.883,0.083)\)