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 .05, what conclusion should be drawn

Answer 1

Answer:

Their welders earn $50,000 annually.

Step-by-step explanation:

Given : $\mu = 50,000$

           $\sigma = 2000$

           n = 100

          $\bar{x}=50350$

To Find : The ship building association wishes to find out whether their welders earn more or less than $50,000 annually.

Solution :

$H_0:\mu = 50000\\H_a: \mu \neq 50000$

n = 100

Since n > 30

So we will use z test

$z= \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Substitute the values

$z= \frac{50350-50000}{\frac{2000}{\sqrt{100}}}$

$z=1.75$

Refer the z table for p value

p value = 0.9599

α= 0.05

Since p value > α

So we accept the null hypothesis .

Hence  their welders earn $50,000 annually.