Answer to Question #146694 in Statistics and Probability for Azz

Question

Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. Is there enough evidence to conclude that the bags do not contain 5 pounds as stated, at α = 0.05? Also, find the 95% confidence interval of the true mean.

Answer 1

Null hypothesis H₀: μ = 5

Alternative hypotheses H₁: μ ≠ 5

Critical value at 0.05 significance level with 49 d.f. z = ±1.96

$z = \frac{4.6 – 5.0}{\frac{0.7}{\sqrt{50}}} = -4.04$

Reject the null hypothesis, since -4.04 < -1.96. There is enough evidence to support the claim that

the bags do not weigh 5 pounds.

The 95% confidence interval for the mean is

$4.6 - (1.96)\frac{0.7}{\sqrt{50}} < μ < 4.6 + (1.96) \frac{0.7}{\sqrt{50}}$

4.4 < μ < 4.8

Notice that the 95 % confidence interval of m does not contain the hypothesized value µ = 5.

Hence, there is agreement between the hypothesis test and the confidence interval.