Question

A pharmaceutical company claims that one of its drugs works on 90% of those who have a certain ailment. An experimenter, suspicious that the company is overexaggerating the drug's efficacy, decides to collect data to test the pharmaceutical company's claim to see if the true efficacy might be less. Of the 250 individuals surveyed, 215 individuals said the drug helped their ailment.

a. What is the null hypothesis?
b. What is the Z-value?
c. Based on an alpha of .10, should we reject or fail to reject the test?
d. What is the result of finding a z-statistic that is more extreme than the critical value?

Answer 1

The null hypothesis is that the drug does not work on 90% of those with the ailment. The Z-value is -1.19. Based on an alpha of 0.10, we fail to reject the null hypothesis. Finding a more extreme Z-statistic than the critical value would provide evidence against the null hypothesis.

Step-by-step explanation:

a. The null hypothesis in this case would be that the drug does not work on 90% of those who have the ailment.

b. To calculate the Z-value, we first need to find the proportion of individuals who said the drug helped their ailment. This can be calculated by dividing the number of individuals who said the drug helped (215) by the total number of individuals surveyed (250). The proportion is 215/250 = 0.86. The Z-value can then be calculated using the formula Z = (observed proportion - expected proportion)/sqrt(expected proportion*(1 - expected proportion)/n), where n is the sample size. In this case, the expected proportion is 0.9, so the Z-value would be (0.86 - 0.9)/sqrt(0.9*(1-0.9)/250) = -1.19.

c. To determine whether we should reject or fail to reject the null hypothesis, we need to compare the Z-value to the critical value. Since the significance level is 0.10, the critical value would be 1.645 for a one-tailed test. Since -1.19 is less than 1.645, we fail to reject the null hypothesis.

d. If we find a Z-statistic that is more extreme than the critical value, it means that the data provides evidence against the null hypothesis. In this case, if we found a Z-statistic greater than 1.645, it would support the alternative hypothesis that the drug's efficacy is less than 90%.

Answer 2

Below is the solution to the given points:

Step-by-step explanation:

Please find the complete question in the attached file.

For point a:

`H_0 : p \geq 0.90`

For point b:

`z=\frac{(x)/(n)-p}{\sqrt{(p(1-p))/(n)}}=\frac{(215)/(250)-0.90}{\sqrt{(0.9 * 0.10 )/(250)}}= -2.11\\\\`

For point c:

`Z_(critical) =z_(\alpha)=-1.28`

since the test statistics
`< Z_(critical)\\\\`

Reject
`H_o`

For point d:

`P(|z \leq -2.11|)=0.0174`