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 3 of 3: construct the $90\%$ confidence interval. round your answers to the nearest integer.

answer
lower endpoint:
upper endpoint:

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 3 of 3: construct the $90\%$ confidence interval. round your answers to the nearest integer.

answer
lower endpoint:
upper endpoint:

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Difference of Means", "Pooled Variance", "Two-Sample t-Interval", "Margin of Error" ], "new_concepts": [], "current_concepts": [ "Difference of Means", "Pooled Variance", "Two-Sample t-Interval", "Margin of Error" ] } ## Calculate the pooled variance and standard deviation $$ \begin{aligned} s_p^2 &= \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \ s_p^2 &= \frac{(11 - 1)(12)^2 + (13 - 1)(17)^2}{11 + 13 - 2} \ s_p^2 &= \frac{10(144) + 12(289)}{22} = \frac{1440 + 3468}{22} = \frac{4908}{22} \approx 223.0909 \ s_p &\approx 14.9362 \end{aligned} $$ ## Determine the critical t-value and margin of error $$ \begin{aligned} df &= n_1 + n_2 - 2 = 22 \ t_{\alpha/2, 22} &\approx 1.717 \quad (\text{for } 90\% \text{ confidence}) \ SE &= s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 14.9362 \sqrt{\frac{1}{11} + \frac{1}{13}} \approx 14.9362 \sqrt{0.1678} \approx 6.1193 \ ME &= t_{\alpha/2} \times SE \approx 1.717 \times 6.1193 \approx 10.5068 \end{aligned} $$ ## Compute the confidence interval endpoints $$ \begin{aligned} \text{Point Estimate} &= \bar{x}_1 - \bar{x}_2 = 90 - 73 = 17 \ \text{Lower Endpoint} &= 17 - 10.5068 = 6.4932 \approx 6 \ \text{Upper Endpoint} &= 17 + 10.5068 = 27.5068 \approx 28 \end{aligned} $$ Construct the 90% confidence interval. Round your answers to the nearest integer. Lower endpoint: 6 Upper endpoint: 28 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Two-Sample t-Interval" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the pooled variance and standard deviation

Determine the critical t-value and margin of error

Compute the confidence interval endpoints

Answer: