Question

The mean annual incomes of certified welders are normally distributed with the mean of $50,000 and a population standard deviation of $2,000. The ship building association wishes to find out whether their welders earn more or less than $50,000 annually. A sample of 100 welders is taken and the mean annual income of the sample is $50,350. If the level of significance is 0.10, what conclusion should be drawn?

A. Do not reject the null hypothesis as the test statistic is less than the critical value of z.
B. Do not reject the null hypothesis as the test statistic is less than the critical value of t.
C. Reject the null hypothesis as the test statistic is greater than the critical value of t.
D. Reject the null hypothesis as the test statistic is greater than the critical value of z.

Answer 1

Answer:

D. Reject the null hypothesis as the test statistic is greater than the critical value of z.

Step-by-step explanation:

$H_{0}:$ welders earn $50,000 annually

$H_{a}:$ welders' income does not equal $50,000 annually

Sample size 100>30, therefore we need to calculate z-values of sample mean and significance.

z-critical at 0.10 significance is 1.65

z-score of sample mean (test statistic) can be calculated as follows:

$\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

• X is the mean annual income of the sample ($50,350)

• M is the mean annual income assumed under null hypothesis ($50,000)

• s is the population standard deviation ($2,000)

• N is the sample size (100)

Then z=$\frac{50,350-50,000}{\frac{2,000}{\sqrt{100} } }$ =

1.75.

Since test statistic is bigger than z-critical, (1.75>1.65), we reject the null hypothesis.