Question

The label on a bag of potato chips claims that there are 3mg of sodium per serving. Suppose an agency is going to test this claim. The agency will only pursue a case if the mean sodium per serving is to high. A sample of 576 bags produced a sample average of 3.700. The standard deviation of sodium per seriving is known to be 6. We want to know whether the new software changed the mean.

1. What is the null hypothesis and what is the alternative hypothesis?

2. What is the value of the test statistic?

3. What is the p-value of the test? (Compute to 4 digits).

4. Do we reject H0 at the .05 level of significance?

5. Do we reject H0 at the .01 level of significance?

Answer 1

Part 1)

To Test :-
H0 :- µ = 3 mg
H1 :- µ ≠ 3 mg

Part 2)

Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 3.7 - 3 ) / ( 6 / √( 576 ))
Z = 2.8

Part 3)

P value = 2 * P ( Z > 2.8 ) = 2 * 1 - P ( Z < 2.8 ) = 0.0051

Part 4)

Reject null hypothesis if P value < α = 0.05 level of significance
Since 0.0051 < 0.05, hence we reject null hypothesis
Result :- Reject null hypothesis

Part 5)

Reject null hypothesis if P value < α = 0.01 level of significance
Since 0.0051 < 0.01, hence we reject null hypothesis
Result :- Reject null hypothesis