Question

A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tests them to make sure that they do not have any dead pixels. Over the course of a month, she tested 300 monitors and found 24 monitors with dead pixels. The technician conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of monitors with dead pixels is greater than 5%.
a. H0:p=0.05; Ha:p>0.05, which is a right-tailed test.
b. Use Excel to test whether the true proportion of monitors with dead pixels is greater than 5%. Identify the test statistic, z, and p-value, rounding to three decimal places.

Answer 1

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis  

$H_{0}: p = 0.05\\H_A: p > 0.05$

This is a one-tailed(right) test.  

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384$

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.