A survey of nonprofit organizations showed that online fundraising has increased in the past year. Based on a random sample of 133 nonprofits, the mean one-time gift donation resulting from email outreach in the past year was $85. Assume that the sample standard deviation is $9. Identify the correct 95% confidence interval estimate for the population mean one-time gift donation.

The answer is below

Step-by-step explanation:

Given that:

Confidence interval (C) = 95%, mean (μ) = 85, standard deviation (σ) = 9, sample size (n) = 133

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.025

The z score of α/2 (0.025) is the same as the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.

The margin of error (E) is given as:

The confidence interval = (μ ± E) = (85 ±  1.53) = (83.47, 86.53)

The confidence interval is between 83.47 and 86.53.

20

step-by-step explanation:

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