Question

Problem 1:

According to the local union president, the mean gross income of plumbers in the Salt Lake City area follows a normal distribution with a mean of $48,000 and a population standard deviation of $2,000. A recent investigative reporter for KYAK TV found, for a sample of 49 plumbers, the mean gross income was $47,600. At the 0.05 significance level, is it reasonable to conclude that the mean income is not equal to $47,600? Determine the p value.

1. State the Null and Alternate hypothesis:
2. State the test statistic:
3. State the Decision Rule:
4. Show the calculation:
5. What is the interpretation of the sample data?
6. Show the P value

Answer 1

It is given that

sample size (n) = 49

population mean () = 48000

sample mean () = 47600

population standard deviation () = 2000

H0 : = 48000

HA : 48000

since we are testing hypothesis for population mean () and population standard deviation () is known then the Test Statistic is given by :

i.e, it follows standard normal distribution

So,

z = (47600-48000)/(2000/7) = -1.4

Looking up the z score table the probability value is 0.08076

Now since this is a two-tailed test hence the p-value is 2*0.08076

i.e p-value = 0.16152

Decision Rule : Reject null-hypothesis if p-value is less than the level of significance.

since the p-value is greater than the level of significance hence we do not reject the null hypothesis at 5% level of significance.

Interpretation of sample data : the mean gross income of plumbers in the Salt Lake City being not equal to $47600 as claimed by the sample data is wrong.