Question

A random sample of 30 boiled peanuts prices are taken. The mean price of the sampled boiled peanuts is $6.75 with a sample standard deviation of $1.50. Find a 90% confidence interval for the mean price of all boiled peanuts. Assume the population is normally distributed.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 6.75

sample standard deviation = s = 1.50

sample size = n = 30

Degrees of freedom = df = n - 1 = 29

t _/2,df = 1.699

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

= 1.699 * (1.50 / 30)

Margin of error = E = 0.47

The 90% confidence interval estimate of the population mean is,

- E < < + E

6.75 - 0.47 < < 6.75 + 0.47

6.28 < < 7.22

(6.28 , 7.22)