Question

A sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 54% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Answer:

So the value of the test statistic is -0.85.

Step-by-step explanation:

We know that a sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 54% of the chips do not fail in the first 1000 hours of their use.

We get that:

$n=1800\\\\p_1=53\%=0.53\\\\p_2=54\%=0.54$

We calculate the standar deviation:

$\sigma=\sqrt{\frac{0.53 \cdot (1-0.53)}{n}}=\sqrt{\frac{0.53 \cdot 0.47}{1800}}=0.01176$

We calculate the value of the test statistic:

$z=\frac{p_1-p_2}{\sigma}=\frac{0.53-0.54}{0.01176}=-0.85$

So the value of the test statistic is -0.85.