Question

A friend of yours attends a different post-secondary institution. She brags that the mean salary of the graduates from her school is over $50000. To test this claim, you collected data from a simple random sample of 32 graduates from her school. Based on the sample data, you found that their mean salary is $51900, with a standard deviation of $5400. Using a 1% significance level, perform a hypothesis test to test your friend’s claim.

Answer 1

Step 1:

Ho: ≤ 50000

Ha: > 50000

Null hypothesis states that the mean salary of graduates from a school is 50000.

Alternative hypothesis states that the mean salary of graduates from a school is greater than 50000.

Step 2: Test statistics

n = 32

sample mean = 51900

sample sd = 5400

Assuming that the data is normally distributed and as the population sd is not given, we will use t stat.

p value

P value = TDIST (t statistics, df, tails) = TDIST(1.99, 31,1) = 0.0277

Step 3

df= 32-1 = 31

level of significance = 0.0

t critical for right tailed test = 2.4528

As the t stat (1.99) does not fall in the rejection area, we fail to reject the Null hypothesis

Also as the p value (0.0277) is greater than level of significance (0.01), we fail to reject the Null hypothesis.

Hence we do not have sufficient evidence to believe that the mean salary of graduates from friend's school is greater than 50000.