Question

An inspector inspects large truckloads of potatoes to determine the proportion p in the shipment with major defects prior to using the potatoes to make potato chips. Unless there is clear evidence that this proportion, p, is less than 0.10, he will reject the shipment. He will test the hypotheses
H0: p = 0.10, Ha: p < 0.10.

e selects an SRS of 100 potatoes from the over 2000 potatoes on the truck. Suppose that six of the potatoes sampled are found to have major defects.
Reference: Ref 18-6

The P-value of this test is
A. 0.4082.
B. 0.0912.
C. 0.0400.
D. less than 0.0002.

A 95% confidence interval for the true proportion of potatoes in the truck that have major defects is
A. 0.034 to 0.086.
B. 0.013 to 0.107.
C. 0.051 to 0.103.
D. 0.026 to 0.128.

Which of the following is true?
A. The inspector will decide to reject the shipment because there's weak evidence that the proportion of potatoes with serious defects is less than 0.10.
B. The inspector might reach the wrong conclusion about the lot of potatoes, whether he returns the shipment or not.
C. Strictly speaking, the inspector should take a larger sample in order to more safely apply the large sample significance test for proportion.
D. All of the above.

Answer 1

P^ = 6/100 = 0.06
P(X < 0.06) = P(z < sqrt(0.06(1 - 0.06)/100)) = P(z < sqrt(0.000564)) = P(z < 0.0237) = 0.5095

95% CI = 0.06 + or - (1.96 x 0.0237) = 0.06 + or - 0.0464 = 0.013 to 0.107

Since the calculated value of z is less than the table value of z, we accept the null hypothesis.

Therefore, The inspector will decide to reject the shipment because there's weak evidence that the proportion of potatoes with serious defects is less than 0.10.