Question

1. The average number of points scored per team in Oregon high school basketball games is μ = 86 with a standard deviation of σ = 7 points. Coach Tom thinks that Portland teams score more than average. Coach Tom

randomly chooses 30 Portland games and finds a sample mean of x = 90 points.

Does this provide sufficient evidence to claim that the population mean of points scored by Portland high school basketball teams is higher than the average of all Oregon high school basketball teams? Use α = 0.05.

1. Step 1: Hypothesize (state the null and alternative hypotheses).

2. Step 2: Prepare (check conditions to apply the Central Limit Theorem).

3. Step 3: Compute (find the value of the test statistic, sketch the appropriate distribution, label both the test statistic and the p-value).

4. Step 4: Interpret (state your conclusion in the context of the application).

Answer 1

Step 1:

Ho: $\mu$ = 86

Ha: $\mu$ > 86

Null hypothesis states that the average number of points scored per team in Portland is equal to 86

Alternative hypothesis states that the average number of points scored per team in Portland is greater than 86

Step 2:

Conditions for central limit theorem:

n = 30 and population sd is given.

Samples are selected randomly & are results are independent

Step 3:

n = 30

sample mean = 90

$z =\frac{ \bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{90 - 86}{7/\sqrt{30}}$

z = 3.130

p value = P(z > 3.130) = 1- P9(z < 3.130) = 1- 0.9991= 0.0009

The z-critical value for a right-tailed test, for a significance level of α=0.05 is zc =1.64

P-Value=0.0009 2 test 23.13 z=7.64

Step 4:

As z stat falls in the rejection area, we reject the Null hypothesis.

Also p value is less than alpha (0.05) we reject the Null hypothesis.

Hence we have sufficient evidence to believe that the average number of points scored per team in Portland is greater than 86