A local five-star restaurant asks its patrons to fill out a quick satisfaction survey (is the patron satisfied or not) after their meal

Question

A local five-star restaurant asks its patrons to fill out a quick satisfaction survey (is the patron satisfied or not) after their meal, and the restaurant claims that their satisfaction rating is 80%. Jennie, a food critic, would like to test this. She takes a sample of 150 patrons, and gets a 78.9% sample satisfaction rating.

(a) Create a 98% confidence interval for the restaurant's true satisfaction rating. (Provide answers in decimal form, not percentage. Thus, 77.3% becomes 0.773. Round your answers to four decimal places, if needed.)

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Answer 1

Hi Kare L.,

In this case, each customer can either report being satisfied or not satisfied. p̂ would be the measured percentage of customers who were satisfied, 78.9% or 0.789.

The formula for a confidence interval would be CI = p̂ ± Z * sqrt((p̂(1-p̂))/n)

Where n is the sample size, in this case 150

and Z is the Z score for your confidence interval, 98%, which you would look up in a z table either online or in your textbook.

This website gives a good explanation of z tables:

https://pro.arcgis.com/en/pro-app/latest/tool-reference/spatial-statistics/what-is-a-z-score-what-is-a-p-value.htm