Question

Pinworm: In Sludge County, a sample of 60 randomly selected citizens were tested for pinworm. Of these, 11 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 0.05 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.
z =

(c) What is the P-value of the test statistic? Use the answer found in the z-table or round to 4 decimal places.
P-value =

(d) What is the critical value of z? Use the answer found in the z-table or round to 3 decimal places.
z_α =

(e) What is the conclusion regarding the null hypothesis?

reject H₀fail to reject H₀    

(f) 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

a)
pcap = 11/60 = 0.183

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.12
Alternative Hypothesis, Ha: p > 0.12

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.183 - 0.12)/sqrt(0.12*(1-0.12)/60)
z = 1.5

c)

P-value Approach
P-value = 0.0668

d)

Rejection Region
This is right tailed test, for α = 0.05
Critical value of z is 1.645

e)

As P-value >= 0.05, fail to reject null hypothesis.

f)

.There is not enough data to support the claim that that the infestation rate in Sludge County is greater than the national average.