Question

2) (3 points) A news report states that the 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (2020, 2160). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 20 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places.

(3 points) A news report states that the 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (2020, 2160). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 20 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places. 1. What is the sample mean? 2. What is the margin of error? 3. What is the sample standard deviation?

Answer 1

The t criticl for df = n - 1 = 20 - 1 = 19 is 1.7291

The Confidence interval is given by $\bar{x}$ $\pm$ ME

The Lower Limit = $\bar{x}$ - ME = 2020 ------- (1)

The Upper Limit = $\bar{x}$ + ME = 2160 -------- (2)

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Adding (1) + (2), we get

2 * $\bar{x}$ = 4180

$\bar{x}$ = Mean = 2090

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Subtracting (2) from (1), we get

2 * ME = 140

Therefore ME = 70

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ME = t critical * Standard Deviation / Sqrt(n)

Therefore SD = (ME * Sqrt(n)) / t critical = (70 * sqrt(20)) / 1.7291

SD = 181.0477

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