Study Guide for Problem #16

Melissa Alteon

Question #16: A survey claims that a college graduate from Smith College can expect an average starting salary of $42,000. Fifteen Smith College graduates, drawn from normally distributed population, had an average starting salary of $40,800 with a standard deviation of $2,250. At the 1% level of significance, can we conclude that the average starting salary of the graduates is significantly less than $42,000?

Step 1: State the claim. Identify the null Hypothesis H0 and alternative hypothesis Ha.

Ho = µ = $42,000

Ha = µ ≠ $42,000

Step 2: Specify the level of significance . The level of significance gives the area of the rejection region(s)

α= .01, because 1% is equivalent to .01

Step 3: Describe the tails of the test. Sketch the rejection region(s)

The test is a two-tailed test since it is NOT equal to $42,000 that only means that it could be either GREATER than $42,000 or LESS than $42,000.

Step 4: Determine the degree of freedom d.f.= n-1

n=15, therefore n-1 would be 15-1 which equals 14

d.f. = 14

Step 5: Determine the critical value(s) using the t-distribution table.

This table is “Table 5- t-Distribution” and how you would determine the critical value is by reading the table left to right. We know that the d.f. = 14 and the test is a two-tailed test and that = .01 so following this method you descend from the value of alpha and you should have gotten

z= 2.977

Step 6: Find the t-value of the test static (from your sample).

z= (x-bar) – µ(x-bar)/ σ(x-bar)

µ (x-bar)= µ, µ= $42,000

σ (x-bar)= , s/ (square root of n)

s= $2250 and n= 15

s/(square root of n) = 580.95

x-bar = $40,80

z= ($40,800-$42,000)/580.95

z=-2.07

Step 7: Make a decision to reject or fail to reject the null hypothesis.

In this case the null hypothesis cannot be rejected because the t-value collected from the sample -2.07 does not lie outside the critical value instead it lies in between the critical values of -2.977 and +2.977. In other words it is not in the rejection area.

Step 8: Interpret the decision in the context of the original claim.

Therefore, A college graduate from Smith College can expect an average starting salary of $42,000 and we can conclude that the average starting salary of the graduates is less than $42,000.

**Step 4: Determine the degree of freedom d.f.= n-1**

n=15, therefore n-1 would be 15-1 which equals 14

d.f. = 14

**Step 5: Determine the critical value(s) using the t-distribution table.**

This table is “Table 5- t-Distribution” and how you would determine the critical value is by reading the table left to right. We know that the d.f. = 14 and the test is a two-tailed test and that = .01 so following this method you descend from the value of alpha and you should have gotten

**z= ** **2.977**

** **

**Step 6: Find the t-value of the test static (from your sample).**

= $42,000

, s= $2250 and n= 15

= 580.95

= $40,800

**Step 7: Make a decision to reject or fail to reject the null hypothesis.**

In this case the null hypothesis **cannot** be rejected because the t-value collected from the sample does not lie outside the critical value instead it lies in between the critical values of -2.977 and +2.977

**Step 8: Interpret the decision in the context of the original claim.**

Therefore, A college graduate from Smith College can expect an average starting salary of $42,000 and we can conclude that the avg starting salary of the graduates is less than $42,000.