Question

Physicians at a clinic gave what they thought were drugs to 860 patients. Although the doctors later learned that the drugs were really placebos,59% of the patients reported an improved condition. Assume that if the placebo is ineffective, the probability of a patients condition improving is 0.55. Test the hypotheses that the proportion of patients improving is >0.55

Find the test statistics:

z=


Find the p-value.
p=

Answer 1

Answer:

Z=2.36

P=0.0091

Step-by-step explanation:

The hypothesis is:

π > 0.55 which is the population proportion

The formula to get the test statistic is

Zstat = P-π/√π(1-π)/n

P =507.4/860 = 0.59 (sample proportion )

n =860 (sample size)

π =0.55(population porportion)

Zstat = 0.59-0.55/√0.55(1-0.55)/860

Zstat =2.36

From Z table the probability of Z score of 2.36 is 0.9909

To calculate for P- value will then be

P= 1 - 0.9909

P= 0.0091

Answer 2

Using the z-distribution, it is found that:

• The test statistic is z = 2.36.
• The p-value of the test is of 0.0091.

The null hypothesis is:

$H_0: p = 0.55$

The alternative hypothesis is:

$H_1: p > 0.55$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are: $p = 0.55, n = 860, \overline{x} = 0.59$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.59 - 0.55}{\sqrt{\frac{0.55(0.45)}{860}}}$

$z = 2.36$

The p-value is the probability of finding a sample proportion of 0.59 or greater, which is 1 subtracted by the p-value of z = 2.36.

• Looking at the z-table, z = 2.36 has a p-value of 0.9909.

1 - 0.9909 = 0.0091.

A similar problem is given at https://brainly.com/question/24166849