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In: Statistics and Probability

Pinworm: In Sludge County, a sample of 40 randomly selected citizens were tested for pinworm. Of...

Pinworm: 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 0.05 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? Round your answer to 4 decimal places.
P-value =

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

reject H0fail 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.

Solutions

Expert Solution

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.12

Ha : p > 0.12

= x / n = 9 / 40 = 0.225

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

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

= 2.04

P-value = 0.0205

= 0.05

P-value <

Reject H0

  We reject the claim that the infestation rate in Sludge County is greater than the national average.

orchestra answered 1 year ago
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