Question

A Study of recent graduates from BSCC revealed that for a sample of 10 recently graduated Accounting Technology majors the mean salary was $32,000 per year with a sample standard deviation of $1,800. A sample of 8 business Administration majors revealed a mean salary of $29,000 per year with a sample standard deviation of $2,000. At the .05 significance level, can we conclude there is a difference in the starting salaries?

1. State the Null and Alternate Hypothesis(H0, H1).

2. Determine the level of significance.

3. Determine the test statistic. (z or t)

4. State the decision rule.(Reject H0 if)

5. Conduct the test and make a decision.(Include Formula and show all work)

6. Interpret the results.

"PLEASE SHOW ALL WORK WITH THE CORRECT ANSWERS"

Thanks!

Answer 1

n1 = 10

= 32000

s1 = 1800

n2 = 8

= 29000

s2 = 2000

1) Claim: There is a difference in the starting salaries.

The null and alternative hypothesis is

2)

Level of significance = 0.05

3)

Here population standard deviations are unknows so we use t-test statistic.

For doing this test first we have to check the two groups have population variances are equal or not.

Null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 2000^2 / 1800^2 = 1.23

Degrees of freedom => n1 - 1 , n2 - 1 => 10 - 1 , 8 - 1 => 9 , 7

Critical value = 3.677   ( Using f table)

Critical value > test statistic so we fail to reject null hypothesis.

Conclusion: The population variances are equal.

So we have to use here pooled variance.

Test statistic is

4)

Degrees of freedom = n1 + n2 - 2 = 10 + 8 - 2 = 16

Critical value = 2.120

( From t table)

If test statistic is greater than the critical value we reject null hypothesis.

5)

Test statistic > critical value we reject null hypothesis.

6)

Interpretation:

There is a difference in the starting salaries.