Critical value: z = 2.33. Fail to reject null hypothesis. There is not sufficient evidence to support the claim
that more than half of all those using the drug experience relief.
-A manufacturer claims that fewer than 6% of its products are defective. In a random sample of 97 such products, 5% are defective. Find the P-value for a test to support the manufacturer's claim.
A) 0.1591
B) 0.3409
C) 0.1736
D) 0.3264

Question

Quizplus · Accepted Answer

We want to test if the true proportion of defective products, p, is less than 6%, i.e., H0: p>= 0.06 versus Ha: p<0.06. 
The test statistic is z = (0.05-0.06)/sqrt(0.06*(1-0.06)/97) = -1.61 (rounded to two decimal places). 
The P-value for a one-tailed test with a test statistic of -1.61 is 0.05, which is larger than the significance level alpha=0.01 (one-tailed), or alpha=0.02 (two-tailed). Therefore, we fail to reject the null hypothesis. This means that there is not enough evidence to support the manufacturer's claim that fewer than 6% of its products are defective at the 1% significance level (one-tailed) or 2% significance level (two-tailed). The P-value is the probability of getting a test statistic as extreme as -1.61, assuming the null hypothesis is true, and it is the probability of getting a z-value less than -1.61 or greater than 1.61 if the distribution were normal. The P-value is calculated using statistical software or a table of the standard normal distribution. In this case, the P-value is approximately 0.3409 (rounded to four decimal places), which is closest to choice B.

Critical Value: Z = 2