Question

Corn: In a random sample of 120 ears of corn, farmer Carl finds that 10 of them have worms. Carl claims that less than 15% of all his corn has worms. Test this claim at the 0.01 significance level.

(a) What is the sample proportion of corn with worms? 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 =   

(e) Choose the appropriate concluding statement.

The data supports the claim that less than 15% of Carl's corn has worms.

There is not enough data to support the claim that less than 15% of Carl's corn has worms.   

We reject the claim that less than 15% of Carl's corn has worms.

We have proven that less than 15% of Carl's corn has worms.

Answer 1

Given nn a random sample of n = 120 ears of corn, farmer Carl finds that X = 10 of them have worms. Carl claims that less than 15%( P = 0.15) of all his corn has worms.

Thus based on the Carl claim the hypotheses are:

Ho : p = 0.15

Ha: p < 0.15

based on the hypothesis it will be a left-tail test.

Normal distribution approximation conditions:

1) The sample must be randomly selected-- Satisfied.

2) n*p(1-p) >=10, so, 120*0.15*(1-0.15) = 15.3 thus satisfied

Since both conditions are satisfied hence we can assume the distribution as normal.

a) The sample proportion is calculated as:

b) The test statistic is calculated as:

Rejection region:

Reject Ho if P-value is less than 0.01.

c) P-value:

The P-value is calculated using the excel formula for normal distribution which is =NORM.S.DIST(-2.045 , TRUE), thus the P-value is computed as:

P-value:0.0204

d) Conclusion:

Since P-value is greater than 0.01 hence we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim,

There is not enough data to support the claim that less than 15% of Carl's corn has worms.