Question

In: Statistics and Probability

Q: The average starting salary of a random sample of 100 high school students was found...

Q: The average starting salary of a random sample of 100 high school students was found to be $31,840. The population standard deviation for all such individuals is known to be $9,840.

a. (12) Ten years ago, the average starting salary was $25,000. Does the sample data support the claim that the starting salary for this group has increased? Use alpha = 0.05.

b. (6) Describe in general Type I and Type II errors and the Power of the test.

c. (6) Describe in the context of the problem Type I and Type II error, and the Power of the test.

please answer ASAP. THANKS!

Solutions

Expert Solution

Solution:

Given:

Sample size = n = 100

Sample mean =

The population standard deviation =

Part a) Ten years ago, the average starting salary was $25,000. thus

Claim: the starting salary for this group has increased

Level of significance =

Step 1) State H0 and H1:

Vs

Step 2) Test statistic:

Step 3) Find z critical value:

Level of significance =

Since this is right tailed test , find Area = 1 -

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Zcritical = 1.645

Step 4)  Decision Rule:
Reject null hypothesis ,if z  test statistic value > z critical value = 1.645, otherwise we fail to reject H0.

Since z  test statistic value = > z critical value = 1.645, we reject null hypothesis H0.

Step 5) Conclusion:

At 0.05 level of significance, the sample data support the claim that the starting salary for this group has increased.

Part b) Describe in general Type I and Type II errors and the Power of the test.

Type I Error = Reject null hypothesis H0, in fact H0 is true.

Type II Error = Fail to reject null hypothesis H0, in fact H0 is False.

Power of the test = 1 - P( Type II Error)

Power of the test = Probability of correctly rejecting null hypothesis when null hypothesis is false.

Part c) Describe in the context of the problem Type I and Type II error, and the Power of the test.

Type I Error = Reject null hypothesis H0, in fact H0 is true.

Type I Error = Conclude that the starting salary for this group has increased ( that is conclude that: the average salary is greater than $25,000) , in fact the starting salary for this group is not greater than $25,000.

Type II Error = Fail to reject null hypothesis H0, in fact H0 is False.

Type II Error = Conclude that: the starting salary for this group is not greater than $25,000, in fact the starting salary for this group has increased ( that is the average salary is greater than $25,000).

Power of the test = Probability of the average salary is greater than $25,000 when it is greater than $25,000.


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