Question

A certain drug carn be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The manufacturer of the drug claims that more than 94% of patients taking the drug are healed within B weeks. In clinical trials, 228 of 240 patients suffering from acid reflux disease were healed after 8 weeks. Test the manufacturer's claim at the a=0.05 level of significance. Click here to view the standard nomal distribution table (page 1). Cick here to view the standard nomal distribution table (page 2). Because rea (1-Po) =| satistied. (Pound to one decimal place as needed.) V 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis

Answer 1

According to the manufacturer's claim, we build an hypothesis test, find the test statistic and the p-value relative to this test statistic, reaching a conclusion that:

The p-value of the test is 0.2578 > 0.05, which means that there is not sufficient evidence to conclude that more than 94% of patients taking the drug are healed within B weeks.

The manufacturer of the drug claims that more than 94% of patients taking the drug are healed within 8 weeks.

At the null hypothesis, we test if the proportion is of at most 0.94, that is:

$H_0: p \leq 0.94$

At the alternative hypothesis, we test if the proportion is of more than 0.94, so:

$H_1: p > 0.94$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.94 is tested at the null hypothesis:

This means that $\mu = 0.94, \sigma = \sqrt{0.94*0.06}$

In clinical trials, 228 of 240 patients suffering from acid reflux disease were healed after 8 weeks.

This means that $n = 240, X = \frac{228}{240} = 0.95$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.95 - 0.94}{\frac{\sqrt{0.94*0.06}}{\sqrt{240}}}$

$z = 0.65$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.95, which is 1 subtracted by the p-value of Z = 0.65.

Looking at the z-table, Z = 0.65 has a p-value of 0.7422.

1 - 0.7422 = 0.2578.

The p-value of the test is 0.2578 > 0.05, which means that there is not sufficient evidence to conclude that more than 94% of patients taking the drug are healed within B weeks.

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