Esteban suspects that right-handed participants have an unfair advantage on a memory test because the input device uses the right hand. He selects a random sample of existing test data and summarizes it in the following table.
\begin{tabular}{lll}
\multicolumn{3}{c}{ Scores on memory test (points) } \
& Left-handed & Right-handed \
\hline Sample mean & $\bar{x}_{\mathrm{L}}=115.3$ & $\bar{x}_{\mathrm{R}}=116.8$ \
Sample standard deviation & $s_{\mathrm{L}}=26.2$ & $s_{\mathrm{R}}=15.8$ \
Sample size & $n_{\mathrm{L}}=9$ & $n_{\mathrm{R}}=24$
\end{tabular}

Assume that the conditions for inference have been met.
Let $\mu_{\mathrm{R}}-\mu_{\mathrm{L}}$ be the difference in mean score of the right- and left-handed participants.

Which of the following is a $90 \%$ confidence interval for $\mu_{\mathrm{R}}-\mu_{\mathrm{L}}$ ?
Use a calculator with statistical capabilities to calculate the interval.
Choose 1 answer:
(A) $(-27.84,30.84)$
(B) $(-19.17,22.17)$
(C) $(-15.33,18.33)$
(D) $(-13.81,16.81)$

Question

Esteban suspects that right-handed participants have an unfair advantage on a memory test because the input device uses the right hand. He selects a random sample of existing test data and summarizes it in the following table.
\begin{tabular}{lll}
\multicolumn{3}{c}{ Scores on memory test (points) } \
& Left-handed & Right-handed \
\hline Sample mean & $\bar{x}_{\mathrm{L}}=115.3$ & $\bar{x}_{\mathrm{R}}=116.8$ \
Sample standard deviation & $s_{\mathrm{L}}=26.2$ & $s_{\mathrm{R}}=15.8$ \
Sample size & $n_{\mathrm{L}}=9$ & $n_{\mathrm{R}}=24$
\end{tabular}

Assume that the conditions for inference have been met.
Let $\mu_{\mathrm{R}}-\mu_{\mathrm{L}}$ be the difference in mean score of the right- and left-handed participants.

Which of the following is a $90 \%$ confidence interval for $\mu_{\mathrm{R}}-\mu_{\mathrm{L}}$ ?
Use a calculator with statistical capabilities to calculate the interval.
Choose 1 answer:
(A) $(-27.84,30.84)$
(B) $(-19.17,22.17)$
(C) $(-15.33,18.33)$
(D) $(-13.81,16.81)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the standard error for the difference in means using the formula $SE = \sqrt{\frac{s_{\mathrm{L}}^2}{n_{\mathrm{L}}} + \frac{s_{\mathrm{R}}^2}{n_{\mathrm{R}}}}$. ‖ ‖ step 2 ⋮ Substitute the given values into the formula: $SE = \sqrt{\frac{26.2^2}{9} + \frac{15.8^2}{24}}$. ‖ ‖ step 3 ⋮ Calculate the value of SE: $SE = \sqrt{\frac{686.44}{9} + \frac{249.64}{24}} = \sqrt{76.27 + 10.40} = \sqrt{86.67} \approx 9.31$. ‖ ‖ step 4 ⋮ Determine the z-score for a 90% confidence interval, which is approximately 1.645 for a two-tailed test. ‖ ‖ step 5 ⋮ Calculate the margin of error (ME) using the formula $ME = z imes SE$. ‖ ‖ step 6 ⋮ Substitute the values into the formula: $ME = 1.645 imes 9.31 \approx 15.34$. ‖ ‖ step 7 ⋮ Calculate the confidence interval using the formula $CI = (\bar{x}_{\mathrm{R}} - \bar{x}_{\mathrm{L}}) \pm ME$. ‖ ‖ step 8 ⋮ Substitute the values into the formula: $CI = (116.8 - 115.3) \pm 15.34 = 1.5 \pm 15.34$. ‖ ‖ step 9 ⋮ Calculate the lower and upper bounds of the confidence interval: Lower bound $= 1.5 - 15.34 = -13.84$, Upper bound $= 1.5 + 15.34 = 16.84$. ‖ ‖ step 10 ⋮ Round the values to two decimal places if necessary: CI $= (-13.84, 16.84)$. ‖ ***D*** ∻Key Concept∻ ⚹Confidence Interval for the Difference in Means⚹ ∻Explanation∻ ⚹The confidence interval for the difference between two means is calculated using the standard error of the difference, the appropriate z-score for the confidence level, and the sample means. The margin of error is then applied to the difference in sample means to find the interval.⚹◊What is the formula for calculating the test statistic in a t-test?⍭ Generate me a similar question◊

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