Question

The potato company sells ”5lb” potato bags. A curious customer wants to check weather the mean weight of ”5lb” potato bag is actually 5lb or not. He collected a sample of 36 ”5 lb” potato bags and found out that the sample mean weight was 5.03 lbs and known population standard deviation was 0.092. Proceed the testing at the significance level α = 0.05. (a) Designate the null and alternative hypotheses. (b) Verify the condition. (c) What is the test statistic? (d) What is the rejection region? (You can calculate p-value instead) (e) State the conclusion in the context of the original problem. (f) Construct 95% confidence interval for µ.

Answer 1

(a)

Null Hypothesis H0: = 5 lb

Altenative Hypothesis Ha:   5 lb

(b)

The sample of 36 ”5 lb” potato bags can be reasonably assumed to be rando,

The sampling distribution is normal or nearly normal as the sample size is greater than 30.

We know the true population standard deviation.

All conditions to perform one sample z test are satisfied.

(c)

Sample mean, = 5.03 lbs

Standard deviation = 0.092 lbs

Standard error of mean, SE = / = 0.092 / = 0.01533

Test statistic, t = ( - ) / SE =  (5.03 - 5) / 0.01533 = 1.957

(d)

For two-tail test, and α = 0.05, rejection region is z < -1.96 or z > 1.96

(e)

Since, the test statistic does not lie in the rejection region, we fail to reject null hypothesis H0 and conclude that there is no significant evidence from the data that weight of ”5lb” potato bag is actually not 5lb

(f)

95% confidence interval for µ is,

(5.03 - 1.96 * 0.01533 ,  5.03 + 1.96 * 0.01533)

(5.00 lbs , 5.06 lbs)