Question

A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 116 residents and found the mean weight to be 177 pounds with a standard deviation of 30 pounds. Determine a 95% confidence interval for the mean, rounding all values to the nearest tenth.

Answer 1

Using a t-distribution with 115 degrees of freedom (df = n-1), and a 95% confidence level, we can find the critical value using a t-table or calculator, which is approximately 1.98.

Then, we can calculate the margin of error (ME) using the formula:

ME = critical value x standard error

where the standard error (SE) is given by:

SE = standard deviation / sqrt(sample size)

Substituting the given values, we get:

SE = 30 / sqrt(116) ≈ 2.78

ME = 1.98 x 2.78 ≈ 5.5

Finally, we can construct the 95% confidence interval (CI) for the mean weight using the formula:

CI = sample mean ± margin of error

Substituting the given values, we get:

CI = 177 ± 5.5

CI ≈ [171.5, 182.5]

Therefore, the 95% confidence interval for the mean weight of the residents in the town is approximately [171.5, 182.5] pounds.

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