Question

A random sample of 31 eight-ounce servings of different juice drinks has a mean of 99.3 calories and a standard deviation of 41.5 calories. Construct a 99% confidence interval for the population mean calorie content of juice drinks. Round all calculation to three decimal places.

a. Explain why the sample size is large enough

b. Find appropriate t value: df=_____ t=_____

c. Calculate the 99% confidence interval.

d. Correctly and completely interpret this confidence interval in context.

Answer 1

a)

Sample size should be large enough (> 30) for approximately normal distribution.

Since n = 31, the distribution is approximately normal.

b)

df = n - 1 = 31 - 1 = 30

From T table,

t critical value at 0.01 significance level with 30 df = 2.750

c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
99.3 ± t(0.01/2, 31 -1) * 41.5/√(31)
Lower Limit = 99.3 - t(0.01/2, 31 -1) 41.5/√(31)
Lower Limit = 78.80
Upper Limit = 99.3 + t(0.01/2, 31 -1) 41.5/√(31)
Upper Limit = 119.80
99% Confidence interval is ( 78.80 , 119.80 )

d)

Interpretation = We are 99% confident that  the population mean calorie content of juice drinks

is between 78.80 calories and 119.80 calories.