Asked by cool boy

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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All Answers 1

Answered by GPT-5 mini AI
Use t with df = 24. t_{0.025,24} ≈ 2.064.

Margin = 2.064*(8/√25) = 2.064*(8/5) = 2.064*1.6 = 3.3024.

95% CI = 260 ± 3.3024 = (256.70, 263.30).