Question

You are given the sample mean and the sample standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals.

A random sample of

4848

​eight-ounce servings of different juice drinks has a mean of

88.688.6

calories and a standard deviation of

43.543.5

calories.

The​ 90% confidence interval is

left parenthesis nothing comma nothing right parenthesis .,.

​(Round to one decimal place as​ needed.)The​ 95% confidence interval is

left parenthesis nothing comma nothing right parenthesis .,.

​(Round to one decimal place as​ needed.)

Which interval is​ wider?

The​ 95% confidence interval

The​ 90% confidence interval

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 88.6

sample standard deviation = s = 43.5

sample size = n = 48

Degrees of freedom = df = n - 1 = 48 - 1 = 47

1) At 90% confidence level the t is,

  = 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

t _/2,df = t _{0.05, 47} = 1.678

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

= 1.678 * ( 43.5 / 48)

Margin of error = E = 10.5

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

- E < < + E

88.6 - 10.5 < < 88.6 + 10.5

78.1 < < 99.1

The 90% confidence interval is,

(78.1,99.1)

2) At 95% confidence level the t is,

  = 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

t _/2,df = t _{0.025, 47} = 2.012

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

= 2.012 * (43.5 / 48)

Margin of error = E = 12.6

The 95% confidence interval estimate of the population mean is,

- E < < + E

88.6 - 12.6 < < 88.6 + 12.6

76.0 < < 101.2

The 95% confidence interval is,

(76.0,101.2)

The 95% confidence interval is wider.