A researcher believes that the percentage of people who exercise in California is greater than the national exercise rate. The national rate is 20%. The researcher gathers a random sample of 120 individuals who live in California and finds that the number who exercise regularly is 31 out of 120.

a. What is Zobt?

b. What is df for this test?

c. What is Chi-sqare cv?

d. What conclusion should be drawn from these results?

If you do a proportional z-test with this data, it would look something like this:

z = (.258 - .20)/√[(.20)(.80)/120]

Note: 31/120 = .258

Calculating:
z = .058/.0365 = 1.589

The null would not be rejected and the conclusion would be that there is no difference in the population of interest.

To answer these questions, we need to perform a hypothesis test using the given data. Here's how you can calculate and interpret the results:

a. Zobt (observed Z-value) represents the number of standard deviations the sample proportion is from the assumed population proportion. To calculate Zobt, use the formula:

Zobt = (P̂ - P) / √((P * (1 - P)) / n)

Where:
P̂ (pronounced "P-hat") is the sample proportion (31/120 = 0.2583)
P is the assumed population proportion (0.20)
n is the sample size (120)

Substituting the values, we get:

Zobt = (0.2583 - 0.20) / √((0.20 * (1 - 0.20)) / 120)
Zobt ≈ 1.3536

b. df (degrees of freedom) for this test is determined by the sample size and the null hypothesis. In this case, since we are comparing a proportion to a known proportion, the formula for df is:

df = n - 1

Substituting the values, we get:

df = 120 - 1
df = 119

c. Chi-square cv (critical value) represents the value from the chi-square distribution that determines the cutoff point for rejecting the null hypothesis. The specific cv depends on the significance level and the degrees of freedom. Since the significance level is not given, let's assume a common value of 0.05. Looking up the chi-square table with 119 df and α = 0.05, we find the critical value to be approximately 90.531.

d. To draw a conclusion from these results, we compare the observed Z-value (Zobt) to the critical value (Chi-square cv). If Zobt is greater than the Chi-square cv, we reject the null hypothesis and conclude that the percentage of people who exercise in California is indeed greater than the national exercise rate.

Since Zobt (1.3536) is less than the Chi-square cv (90.531), we do not have enough evidence to reject the null hypothesis. Therefore, we cannot conclude that the percentage of people who exercise in California is greater than the national exercise rate at the chosen significance level.