Question

construct a 90% confidence interval for $\mu_1 - \mu_2$. two samples are randomly sel sample statistics are given below. assume that $\sigma_1^2 = \sigma_2^2$. $n_1 = 10, n_2 = 12, \bar{x}_1 = 25, \bar{x}_2 = 23, s_1 = 1.5, s_2 = 1.9$ a. $(1.413, 3.124)$ b. $(0.721, 3.279)$ c. $(1.335, 3.012)$ d. $(1.554, 3.651)$

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 \( n_1 = 10 \), \( n_2 = 12 \), \( s_1 = 1.5 \), \( s_2 = 1.9 \):
\( (n_1 - 1)s_1^2=(10 - 1)\times1.5^2 = 9\times2.25 = 20.25 \)
\( (n_2 - 1)s_2^2=(12 - 1)\times1.9^2 = 11\times3.61 = 39.71 \)
\( n_1 + n_2-2=10 + 12-2 = 20 \)
\( s_p^2=\frac{20.25 + 39.71}{20}=\frac{59.96}{20}=2.998 \)

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 \( s_p=\sqrt{s_p^2}=\sqrt{2.998}\approx1.731 \)
\( \frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{10}+\frac{1}{12}=\frac{6 + 5}{60}=\frac{11}{60}\approx0.1833 \)
\( SE = 1.731\times\sqrt{0.1833}\approx1.731\times0.428\approx0.741 \)

Step3: Find t - critical value

The degrees of freedom \( df=n_1 + n_2-2 = 20 \), and for a 90% confidence interval, \( \alpha = 0.10 \), so \( \alpha/2=0.05 \).
From t - table, \( t_{0.05,20}\approx1.725 \)

Step4: Calculate margin of error (ME)

\( ME=t_{\alpha/2,df}\times SE=1.725\times0.741\approx1.278 \)

Step5: Calculate difference in sample means

\( \bar{x}_1-\bar{x}_2=25 - 23 = 2 \)

Step6: Calculate confidence interval

The confidence interval for \( \mu_1-\mu_2 \) is \( (\bar{x}_1-\bar{x}_2 - ME,\bar{x}_1-\bar{x}_2 + ME) \)
Lower bound: \( 2-1.278 = 0.722 \)
Upper bound: \( 2 + 1.278 = 3.278 \)
The interval \( (0.721,3.279) \) is the closest (due to rounding differences in calculations).

Answer:

B. (0.721,3.279)