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A sample survey finds that
30% of a sample of 874 Ohio adults said good health was the thing they were most thankful for. If that sample were an SRS from the population of all Ohio adults, what would be the standard deviation for a
99% confidence interval for the percent of all Ohio adults who feel that way?

$2$ . A sample survey finds that $30 \%$ of a sample of $874$ Ohio adults said good health was the thing they were most thankful for. If that sample were an SRS from the population of all Ohio adults, what would be the standard deviation for a $99 \%$ confidence interval for the percent of all Ohio adults who feel that way?

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Interpret confidence intervals for population means

Full solution

Q.

2

. A sample survey finds that

30 \%

of a sample of

874

Ohio adults said good health was the thing they were most thankful for. If that sample were an SRS from the population of all Ohio adults, what would be the standard deviation for a

99 \%

confidence interval for the percent of all Ohio adults who feel that way?

Calculate sample proportion: Calculate the sample proportion ( $\hat{p}$ ). $\newline$ $\hat{p} = 30\% = 0.30$
Find number of adults: Find the number of adults in the sample $n$ . $n = 874$
Calculate standard error: Calculate the standard error (SE) using the formula $SE = \sqrt{\hat{p}(1 - \hat{p}) / n}$ . $SE = \sqrt{0.30 \times (1 - 0.30) / 874}$ $SE = \sqrt{0.21 / 874}$ $SE = \sqrt{0.000240274}$ $SE \approx 0.0155$
Determine z-score: Determine the z-score for a $99\%$ confidence interval. $\newline$ For a $99\%$ confidence interval, the z-score is approximately $2.576$ .
Calculate standard deviation: Calculate the standard deviation for the $99$ % confidence interval using the formula $SD = z \times SE$ . $\newline$ $SD = 2.576 \times 0.0155$ $\newline$ $SD \approx 0.03993$

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