Question

A company that produces 8-oz low-fat yogurt cups wanted to estimate the mean number of calories for such cups. A random sample of 10 such cups produced the following numbers of calories: Assume that the number of calories in such cups are normally distributed.

147 159 153 146 144 148 163 153 143 158

a. What is the point estimate of the mean calories for 8-oz low-fat yogurt cups?

b. Construct a 95% confidence interval for the mean calories for this sample.

c. For a significance level of 5%, test the statement that average number of calories in 8-oz low-fat yogurt cups is less than 160 calories.

Answer 1

a)

Point estimate = sample mean, xbar = 151.4

b)

sample standard deviation, s = 6.8832
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262

ME = tc * s/sqrt(n)
ME = 2.262 * 6.8832/sqrt(10)
ME = 4.924

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (151.4 - 2.262 * 6.8832/sqrt(10) , 151.4 + 2.262 * 6.8832/sqrt(10))
CI = (146.48 , 156.32)

C)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 160
Alternative Hypothesis, Ha: μ < 160

Rejection Region
This is left tailed test, for α = 0.05 and df = 9
Critical value of t is -1.833.
Hence reject H0 if t < -1.833

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (151.4 - 160)/(6.8832/sqrt(10))
t = -3.951

P-value Approach
P-value = 0.0017
As P-value < 0.05, reject the null hypothesis.