Question

A certain drug is used to treat asthma. In a clinical trial of the​ drug, 23 of 277 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts​ (a) through​ (e) below.

a. What is test statistic?
b. What is p-value?
c. What is the null hypothesis, and what can we conclude about it?
d. Decide whether to reject the null hypothesis?
e. What is the final conclusion?

Answer 1

Answer:

a) The test statistic is z=-0.94

b) The p-value is 0.1736

c) The null hypothesis is $H_0:p=0.10$, we can conclude that, if the result is not significant at 0.01 level, we fail to reject the null hypothesis

d) We fail to reject the null hypothesis at 0.01 significance level.

e) We do not have sufficient  evidence to reject the claim that, less than 10​% of the treated subjects experienced headaches.

Step-by-step explanation:

The test statistic is defined by:

$Z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n} } }$

It was given that, 23 of 277 treated in the clinical trial of the​ drug subjects experienced headaches.

This means that:

$\hat p=\frac{23}{277}\approx 0.083$  and n=277

The  claim is that, less than 10​% of treated subjects experienced headaches. This means:

$p_0=\frac{10}{100}=0.1$

We substitute the values into the formula to obtain:

$Z=\frac{0.083-0.1}{\sqrt{\frac{0.1(1-0.1)}{277} } }$

The test statistics becomes:

$Z=-0.94$

b) From the normal distribution table, the p-value corresponds to Z=-0.94 .

Since this is a left-tailed test, the p-value corresponds to area under the normal curve that is to the left of z=-0.94

P(Z<-0.94)=0.173609.

c) Since the claim is that, less than 10​% of treated subjects experienced headaches, the null hypothesis is

$H_0:p=0.10$

The alternate hypothesis will be:

$H_1:p\:<\:0.10$

This implies that, the result is not significant at 0.01 alpha level.

d) We need to compare the p-value to the significance level. If the significance level is greater than the null hypothesis, we reject the null hypothesis.

Since 0.01<0.1736, we fail to reject the null hypothesis.

d) Conclusion: There is no enough evidence to reject the claim that, less than 10​% of treated subjects experienced headaches.

Answer 2

Answer:

(a)Test statistic$z_{score}=-0.94$

(b) p-value=0.1736

(c)Null hypothesis,

    $H_0:p=0.10$

(d)$\alpha>p$, we reject the null hypothesis.

(e)We have experimental values to reject the claim.

Step-by-step explanation:

Given information;

n=277

Significance level $\alpha$=0.01

By normal distribution,

$z_{score}=\frac{\widehat{p}-p_0}{\frac{\sqrt {p_0(1-p_0)}}{n}}$

$\widehat{p}=\frac{23}{277}=0.83$

$p_0=10$%=0.1

On substitution,

$z_{score}=\frac{{0.83}-0.1}{\frac{\sqrt {0.1(1-0.1)}}{277}}$

Test static,

$z_{score}=-0.94$

b)From the table normal distribution,

for,$z_{score}=-0.94,$

$P(z<-0.94)=0.173609$

(c)Null hypothesis,

    $H_0:p=0.10$

     Alternate hypothesis,

    $H_1:p<0.10$

It implies result is not significant at

$\alpha=0.01$

(d) On compare value if

$\alpha>p$, we reject the null hypothesis.

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