Question

Based on a random sample of 20 people, the mean tip that they said they would leave after a $30 meal was $5.00 and the standard deviation was $0.25. Assume the tip amounts are normally distributed. Construct a 99% confidence interval for the population mean tip. a) Identify the sample size, sample mean, sample standard deviation, and the appropriate test score (t or z) value for 99% confidence. b) Calculate the standard error of the mean. c) Calculate the maximum error of the mean. d) Determine the confidence interval. e) Explain what the confidence interval means in the words of this problem.

Answer 1

Solution :

Point estimate = sample mean = = 5.00

sample standard deviation = s = 0.25

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 -1 = 19

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

t/2,df = t0.005,19 = 2.86

t/2,df = 2.86

Standard error = (s /n) = (0.25 / 20) = 0.0560

Margin of error = E = t/2,df * (s /n)

= 2.86 * (0.25 / 20)

Margin of error = E = 0.16

The 99% confidence interval estimate of the population mean is,

- E < <  + E

5 - 0.16 < < 5. +0.16

4.84 < < 5.16

(4.84 ,5.16)

A confidence interval is an interval estimate combined with a probability statement.

Confidence interval, in statistics, refers to the probability that a population.