Question

The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meal per week. Construct a 99% confidence interval for the population mean.

• Explain which words in the problem indicate a z-distribution or a t-distribution (choose one).
• Show how to calculate the confidence interval using Excel or typing out the equation (such as for proportions). The purpose is to identify a correct equation (with numbers typed out for all to learn) to solve these problems. Note: You must pick your own problem for this post that has not yet been used by another student. Put the problem number you chose in the title of your post

Answer 1

Since the population standard deviation is not known, but sample satandard deviation s known, we can use t distribution.

The sample size = 60 so degrees of freedom = 59. For 99% confidence interval, confidence interval is 1%.

Hence t value from t-table is 2.66.

Sample mean = 2.76 and sample standard deviation s = 0.75.

X

Hence, 99% Confidence interval is $(\bar X - t*s/\sqrt{n}, \bar X + t*s)/\sqrt{n})$

=

(2.76 – 2.66 * 0.75/V60, 2.76 +2.66 *0.75/V60)

= (2.50, 3.02)