An SRS of 400 high school seniors gained an average of i = 21.87 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation o = 48.66. We want to estimate the mean change in score µ in the population of all high school seniors. (a) Using the 68–95–99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for µ based on this sample. (Enter your answers rounded to three decimal places. If you are using CrunchIt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.)

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An SRS of 400 high school seniors gained an average of \( \bar{x} = 21.87 \) points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation \( \sigma = 48.66 \). We want to estimate the mean change in score \( \mu \) in the population of all high school seniors.

(a) Using the 68–95–99.7 Rule or the z-table (Table A), give a 95% confidence interval \( (a, b) \) for \( \mu \) based on this sample.

(Enter your answers rounded to three decimal places. If you are using CrunchIt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.)

\[ a: \]
\[ b: \]

(b) Based on your confidence interval in part (a), how certain are you that the mean change in score \( \mu \) in the population of all high school seniors is greater than 0?

- The upper endpoint of the interval is larger than 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0.

- We cannot be certain at all. If we want to make conclusions about the population mean change in score, we should do a hypothesis test instead of a confidence interval.

- The entire interval is larger than 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0.

- The interval does not contain 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0.
Transcribed Image Text:An SRS of 400 high school seniors gained an average of \( \bar{x} = 21.87 \) points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation \( \sigma = 48.66 \). We want to estimate the mean change in score \( \mu \) in the population of all high school seniors. (a) Using the 68–95–99.7 Rule or the z-table (Table A), give a 95% confidence interval \( (a, b) \) for \( \mu \) based on this sample. (Enter your answers rounded to three decimal places. If you are using CrunchIt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.) \[ a: \] \[ b: \] (b) Based on your confidence interval in part (a), how certain are you that the mean change in score \( \mu \) in the population of all high school seniors is greater than 0? - The upper endpoint of the interval is larger than 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0. - We cannot be certain at all. If we want to make conclusions about the population mean change in score, we should do a hypothesis test instead of a confidence interval. - The entire interval is larger than 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0. - The interval does not contain 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0.
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