Question

In Sludge County, a sample of 40 randomly selected citizens were tested for pinworm. Of these, 9 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.01 significance level.

a) What is the sample proportion of Sludge County residents with pinworm? Round your answer to 3 decimal places. p̂ =

(b) What is the test statistic? Round your answer to 2 decimal places. zp hat =

(c) What is the P-value of the test statistic? Round your answer to 4 decimal places. P-value =

(d) What is the conclusion regarding the null hypothesis?
- reject H0
-fail to reject H0

(e) Choose the appropriate concluding statement.
-The data supports the claim that the infestation rate in Sludge County is greater than the national average.
-There is not enough data to support the claim that that the infestation rate in Sludge County is greater than the national average.
-We reject the claim that the infestation rate in Sludge County is greater than the national average.
-We have proven that the infestation rate in Sludge County is greater than the national average.

Answer 1

Solution :

Given that ,

n = 40

x = 9

The null and alternative hypothesis is

H₀ : p = 0.12

H_a : p > 0.12

This is the right tailed test .

= x / n = 9 / 40 = 0.225

(a) = 0.225

P₀ =  12% = 0.12

1 - P₀ = 1 - 0.12 = 0.88

(b) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.225 - 0.12 / [0.12 * (0.88) / 40]

= 2.04

Test statistic = 2.04

(c) P-value = 0.0207

= 0.01

0.0207 > 0.01

P-value >

(d) Fail to reject Ho

(e) There is not enough data to support the claim that the infestation rate in Sludge Country is greater than the National average.