Question

A random sample of 25 office workers was selected to determine how much the average person spends weekly on groceries. The sample mean was equal to $260, with a sample standard deviation of $8. What is the 95 percent confidence interval of this sample? Round the answer to the nearest hundredth.(1 point)
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Incorrect

Use the margin of error to determine the lower bound of the 95 percent confidence interval.
Use the margin of error to determine the upper bound of the 95 percent confidence interval.
Remaining Attempts : 2

Answer 1

Degrees of freedom = 24. t_{0.975,24} ≈ 2.0639.

Margin of error = 2.0639*(8/√25) = 2.0639*(8/5) = 2.0639*1.6 ≈ 3.30224.

95% CI = 260 ± 3.30224 → (256.70, 263.30) (rounded to nearest hundredth).