Answer to Question #236968 in Statistics and Probability for blossomqt

Question

The advertised claim for batteries for cell phones is set at 48 operating hours with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours. Do these experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, with proper charging procedures? Use a hypothesis-testing procedure with α = 0.01.

Answer 1

Let p denotes the population proportion.

$H_0: p =0.002 \\ H_1: p< 0.002$

Let $\hat{p}$ denotes the sample proportion and n denotes the sample size.

$\hat{p}= \frac{15}{5000}= 0.003 \\ n=5000 \\ Standard \;error = \frac{0.002 \times (1-0.002)}{5000}= 0.000632$

Test-statistic:

$Z = \frac{\hat{p} -0.002}{\sqrt{\frac{0.002 \times (1-0.002)}{n}}} \\ Z = \frac{0.003-0.002}{\sqrt{\frac{0.002(1-0.002)}{5000}}} = 1.5827$

Decision rule:

Reject H₀ at 0.01 level of significance if p-value < 0.01 or if $Z< -Z_{0.01} = -2.3263$

P-value = P(Z<1.5827) = 0.9432

Since p-value > 0.01 and Z > -2.3263, we fail to reject H0 at 0.01 level of significance.

We conclude that the population proportion is not significantly less than 0.002.