Question

I. A university wants to start a hockey team finding that 83% of a random sample of 100 people support the establishment of said team. However, the sample appears to contain 80 mean and 20 women; another random sample of the university body suggests that the true percentage of support for a hockey team is 55%. Calculate the test statistic one would use to determine if there is a significant difference between the male dominated sample and the "normal" sample.

II. For the given significance test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified. A health insurer has determined that the "reasonable and customary" fee for a certain medical procedure is $1200. They suspect that the average fee charged by one particular clinic for this procedure is higher than $1200. The insurer performs a significance test to determine whether their suspicion is correct using α = 0.05. The hypotheses are:
H0: μ = $1200
Ha: μ > $1200
If the P-value is 0.09 and a decision error is made, what type of error is it? Explain.

a) Type II error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher.
b) Type I error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher.
c) Type II error. We conclude that the average fee charged for the procedure is higher than $1200 when it actually is not higher.
d) Type I error. We conclude that the average fee charged for the procedure is higher than $1200 when it actually is not higher.

Answer 1

Solution:-

I)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.55
Alternative hypothesis: P 0.55

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = 0.04975
z = (p - P) /S.D

z = 5.63

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 5.63 or greater than 5.63.

P-value = P(z < - 5.63) + P(z > 5.63)

Use z-calculator to find the p-values.

P-value = 0.00 + 0.00

Thus, the P-value = 0.00

Interpret results. Since the P-value (0.00) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significant difference between the male dominated sample and the "normal" sample.

II) a) Type II error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher.

Type I error:

In statistical hypothesis testing, a type I error is the rejection of a true null hypothesis.

The type II error is the non-rejection of a false null hypothesis.

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