Question

A sample of 1500 computer chips revealed that 62% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 63% 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.
a) Is there enough evidence at the 0.10 level to support the manager's claim?

Answer 1

Answer:

We conclude that the actual percentage that do not fail is equal to the stated percentage which means the manager's claim is not supported.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 62% of the chips do not fail in the first 1000 hours of their use.

The company's promotional literature states that 63% of the chips do not fail in the first 1000 hours of their use.

Let p = actual percentage of computer chips that do not fail

So, Null Hypothesis, $H_0$ : p = 63%     {means that in the actual percentage that do not fail is equal to the stated percentage}

Alternate Hypothesis, $H_A$ : p $\neq$ 63%     {means that in the actual percentage that do not fail is different from the stated percentage}

The test statistics that would be used here One-sample z test for proportions;

                             T.S. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample % of the computer chips that do not fail = 62%

            n = sample of computer chips = 1500

So, the test statistics  =  $\frac{0.62-0.63}{\sqrt{\frac{0.62(1-0.62)}{1500} } }$  

                                       =  -0.798

The value of z test statistics is -0.798.

Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage that do not fail is equal to the stated percentage.