Question

The national average of college students on a test of sports trivia is 50 with a standard deviation of 5. A sportscaster is interested in whether BC students know less about sports than the national average. The sportscaster tests a random sample of 25 BC students and obtains a mean of 48 Use an alpha level of 0.05.

1. State the z score(s) that form the boundaries of the critical region. Use an alpha level of 0.05.

2. Calculate the standard error

3. Calculate the z score

4. What decision would you make? Fail to reject the null hypothesis or Reject the null hypothesis

5. What can the researcher conclude? Let’s say in the above example that the BC average on the sports trivia test was 52 resulting in a z score of +2.0. What decision would the researcher make? Fail to reject the null hypothesis? or reject the null hypothesis?

6. Why did you make that decision in Question 5?

Answer 1

Step 1:

Ho:

Ha:

Null hypothesis states that the BC students average on a test of sports trivia is 50

Step 2: Test statistics

n = 25

sample mean = 48

= 5

Assuming that the data is normally distributed. Also as the population SD is given, we will use z stat

(1) z score and critical region

The z-critical value for a left-tailed test, for a significance level of α=0.05

zc=−1.64

(2) Standard error

(3) z score = -2.00

(4) As the z stat (-2.00) falls in the rejection area, we reject the Null hypothesis.

(5) As the Null hypothesis is rejected, the researches conclude that there sufficient evidence to believe that the BC student's average on a test of sports trivia is less than 50

In case the average is 52 and z value is 2.0, we fail to reject the Null hypothesis.

(6) as the z score 2.00 does not fall in the rejection area. z critical is -1.64