given two independent random samples with the following results:
\
\begin{aligned}
n_1 &= 11 & n_2 &= 13 \\
\bar{x}_1 &= 90 & \bar{x}_2 &= 73 \\
s_1 &= 12 & s_2 &= 17
\end{aligned}
\
use this data to find the $90\%$ confidence interval for the true difference between the population means. assume that the population variances are equal and that the two populations are normally distributed.

step 2 of 3: find the margin of error to be used in constructing the confidence interval. round your answer to six decimal places.

Question

given two independent random samples with the following results:
\
\begin{aligned}
n_1 &= 11 & n_2 &= 13 \\
\bar{x}_1 &= 90 & \bar{x}_2 &= 73 \\
s_1 &= 12 & s_2 &= 17
\end{aligned}
\
use this data to find the $90\%$ confidence interval for the true difference between the population means. assume that the population variances are equal and that the two populations are normally distributed.

step 2 of 3: find the margin of error to be used in constructing the confidence interval. round your answer to six decimal places.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Pooled Variance", "Two-Sample t-Interval", "Margin of Error" ], "current_concepts": [ "Pooled Variance", "Two-Sample t-Interval", "Margin of Error" ] } ## Calculate the pooled sample standard deviation $$ \begin{aligned} s_p^2 &= \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \ s_p^2 &= \frac{(11 - 1)(12)^2 + (13 - 1)(17)^2}{11 + 13 - 2} \ s_p^2 &= \frac{10(144) + 12(289)}{22} \ s_p^2 &= \frac{1440 + 3468}{22} = \frac{4908}{22} \approx 223.090909 \ s_p &\approx 14.936228 \end{aligned} $$ ## Determine the critical value $$ \begin{aligned} df &= n_1 + n_2 - 2 = 22 \ \alpha &= 1 - 0.90 = 0.10 \ t_{\alpha/2, df} &= t_{0.05, 22} \approx 1.717144 \end{aligned} $$ ## Calculate the margin of error $$ \begin{aligned} E &= t_{\alpha/2, df} \cdot s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \ E &= 1.717144 \cdot 14.936228 \cdot \sqrt{\frac{1}{11} + \frac{1}{13}} \ E &\approx 25.647656 \cdot \sqrt{0.090909 + 0.076923} \ E &\approx 25.647656 \cdot \sqrt{0.167832} \ E &\approx 25.647656 \cdot 0.409673 \approx 10.507151 \end{aligned} $$ Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. 10.507151 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Margin of Error" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the pooled sample standard deviation

Determine the critical value

Calculate the margin of error

Answer: