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.