Question

The Acme Battery Company has developed a new cell phone battery. On average, the battery lasts 60 minutes on a single charge. The standard deviation is 4 minutes. Suppose the manufacturing department runs a quality control test. They randomly select 7 batteries. The standard deviation of the selected batteries is 6 minutes. a. What would be the chi-square statistic represented by this test? b. What are the degrees of freedom for the chi-square statistic? c. Suppose they repeated the test with a new random sample of 7 batteries. What is the probability that the standard deviation in the new test would be greater than 6 minutes? (Use the applet)

Answer 1

X² = 13.5

Degree of freedom = 6

0.04

Explanation:

Give the following :

Number of samples (n) = 7

Population standard deviation (σ) =.4 minutes

Sample standard deviation (s) = 6

Mean battery life = 60 minutes

A) Chisquare (X²) :

X² = ((n - 1) * s²) / σ²

X² = ((7 - 1) * 6²) / 4²

X² = (6 * 6²) / 4²

X² = (6 * 36) / 16

X² = 216 / 16

X² = 13.5

B.) The degree of freedom:

degree of freedom = n - 1

Degree of freedom = 7 - 1 = 6

C.) using a chisquare distribution calculator :

With degree of freedom (df) = 6

X² = 13.5

Hence,

P(X² > 13.5) = 0.04 ( from calculator)