Question

He percentage of defective chips produced each hour by oven A has a mean of 33. 56 and a standard deviation of 5. 20. The percentage of defective chips produced each hour by oven B has a mean of 32. 44 and a standard deviation of 3. 78. The hourly differences in percentages for oven A minus oven B have a mean of 1. 11 and a standard deviation of 4. 28. Does there appear to be a difference between oven A and oven B with respect to the mean percentages of defective chips produced

Answer 1

Using the z-distribution, as we have the standard deviation for the population, it is found that it does not appear that there is a difference.

What are the hypothesis tested?

At the null hypothesis, it is tested that there is no difference, that is, the mean of the distribution of the differences is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if there is a difference, hence:

$H_1: \mu \neq 0$

What is the test statistic?

It is given by:

$z = \frac{\overline{x} - \mu}{s}$

In which:

• $\overline{x}$ is the mean of the distribution of differences.

• $\mu = 0$ is the value tested at the null hypothesis.

• s is the standard error.

In this problem, the parameters are as follows: $\mu = 1.11, s = 4.28$.

Hence:

$z = \frac{\overline{x} - \mu}{s}$

$z = \frac{1.11 - 0}{4.28}$

$z = 0.26$

What is the decision?

Considering a two-tailed test, as we are testing if the mean is different of a value, with a standard significance level of 0.05, the critical value is of $|z^{\ast}| = 1.96$.

Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis, that is, it does not appear that there is a difference.

To learn more about the z-distribution, you can check https://brainly.com/question/26454209