The filling variance for boxes of cereal is designed to be .02 or less. A sample of 41 boxes of cereal shows a sample standard deviation of 0.16 ounces. a. Use α= 0.05 to determine whether the variance in the cereal box fillings is exceeding the design specification.
b. State the null and alternative hypotheses.
c. Calculate the value of the test statistic (to 2 decimals).
d. The p-value is:
e. What is your conclusion?

-

Null hypothesis             H₀ : σ² ≤ 0.02

Alternative hypothesis Hₐ  : σ² > 0.02

- Chi Square = 51.2

- p-value = 0.1104

- since the calculated p-value ( 0.1104 ) is greater than the level of significance ( 0.05 ).

We fail to reject null hypothesis.

- So at, 5% level of significance, there is no evidence to support that the variance in the cereal box fillings is designed to be 0.02 or less.

Step-by-step explanation:

Given the data in the question;

filling variance for boxes of cereal is designed to be .02 or less

sample size n = 41 boxes

standard deviation S = 0.16 ounces

level of significance = 0.05

Hypothesis;

Null hypothesis             H₀ : σ² ≤ 0.02

Alternative hypothesis Hₐ  : σ² > 0.02

{ Upper tail test }

Test statistics;

Chi Square = [ ( n-1 ) × S² ] / σ²

Chi Square = [ ( 41-1 ) × (0.16)² ] / 0.02

Chi Square = [ 40 × 0.0256 ] / 0.02

Chi Square = 1.024/ 0.02

Chi Square = 51.2

Degree of freedom df = n - 1 = 41 - 1 = 40

Chi square value at 0.05  and df of 40 = 55.7585

p-value = 0.1104

since the calculated p-value ( 0.1104 ) is greater than the level of significance ( 0.05 ).

We fail to reject null hypothesis.

So at, 5% level of significance, there is no evidence to support that the variance in the cereal box fillings is designed to be 0.02 or less.

80% of the cookies

step-by-step explanation: