Question

Suppose a university advertises that its average class size is 32 or less. A student organization is concerned that budget cuts have led to increased class sizes and would like to test this claim. A random sample of 39 classes was​ selected, and the average class size was found to be 34.7 students. Assume that the standard deviation for class size at the college is 7 students. Using α=0.01​, complete parts a and b below.

a. Does the student organization have enough evidence to refute the​ college's claim?

Determine the null and alternative hypotheses.

H0​: μ is greater than > less than or equals ≤ greater than or equals ≥ not equals ≠ less than < equals = n

H1​:μ ▼ greater than or equals ≥ not equals ≠ equals = less than < greater than > less than or equals ≤

The​ z-test statistic is . ​(Round to two decimal places as​ needed.)

The critical​ z-score(s) is(are) . ​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Because the test statistic ▼ is greater than the critical value/is less than the critical value/falls within the critical values/does not fall within the critical values, reject/do not reject the null hypothesis.

b. Determine the​ p-value for this test.

The​ p-value is . ​(Round to three decimal places as​ needed.)

Answer 1

Step 1:

Ho:

Ha:

Step 2:

n = 39

sample mean = 34.7

population standard deviation = 7

Assuming that the data is normally distrbuted and population sd is given, we will calculate z statistics

z statistics = 2.41

Step 3:

The z-critical value for a right-tailed test, for a significance level of α=0.01

zc = 2.33

As the z stat ( 2.41) is greater than z critical (2.33) i.e. falls in the rejection area, we reject the Null

hypothesis.

(b) P value

P (z > 2.41) =

P ( z > 2.41 )=1−P ( z < 2.41 )=1− 0.992 = 0.008

As the p value (0.008) is less than level of significance (0.01) we reject the Null hypothesis.