91影视

A sample of 150 patients was treated with oxygen, and among them, 116 were free from headaches 15 minutes after treatment. Another sample of 148 patients was given a placebo, and among them, 29 were free from headaches 15 minutes after treatment. The significance level is $伪=0.01$.

It is claimed that oxygen treatment is effective; that is, the proportion of patients free from headache after oxygen treatment is greater than the proportion of patients free from headache after receiving a placebo.

Since the given claim does not have an equality sign, the following hypotheses are set up:

Null Hypothesis: The proportion of patients who are free from headache 15 minutes after oxygen treatment is equal to theproportion of patients who are free from headache 15 minutes after receiving placebo.

$H_0:p_1=p_2$

$H_1:p_1>p_2$

The test is right-tailed.

Let ${\hat{p}}_1$denote the sampleproportion of patients free from headache 15 minutes after oxygen treatment.

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{116}{150} \\ =0.7733 \end{gathered}$$

Let ${\hat{p}}_2$denote the sampleproportionof patients free from headache 15 minutes after receiving placebo.

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{29}{148} \\ =0.1959 \end{gathered}$$

The sample size of patients who weretreated with oxygen treatment $\left(n_1\right)$is equal to 150.

The sample size of patients who weretreated with a placebo$\left(n_2\right)$ is equal to 148.

The value of the pooled sample proportion is computed as follows:

$$\begin{gathered} \stackrel{炉}{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)}鈥� \\ =\frac{\left(116+29\right)}{\left(150+148\right)} \\ =0.4866 \end{gathered}$$

and

$$\begin{gathered} \stackrel{炉}{q}=1-\stackrel{炉}{p} \\ =1-0.4866 \\ =0.5134 \end{gathered}$$

The test statistic is computed as follows:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_1}+\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_2}\right)}} \\ =\frac{\left(0.7733-0.1959\right)-0}{\sqrt{\left(\frac{0.4866脳0.5134}{150}+\frac{0.4866脳0.5134}{148}\right)}} \\ =9.971 \end{gathered}$$

The value of the test statistic is 9.971.

Referring to the standard normal distribution table, the critical value of z corresponding to$伪=0.01$ for a right-tailed test is equal to 2.33.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0001.

Since the p-value is less than 0.01, the null hypothesis is rejected.

a.

There is sufficient evidence to support the claim that theproportion of patients who were free from headache 15 minutes after oxygen treatment is greater than the proportion of patients who were free from headache 15 minutes after placebo treatment.

b.

The general formula for the confidence interval of difference of proportion is written below:

$\text{Confidence}鈥�鈥�\text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E鈥�鈥�,鈥�鈥�\left({\hat{p}}_1-{\hat{p}}_2\right)+E\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

Where, E is the margin of error and has the following formula:

$E=z_{\frac{伪}{2}}脳\sqrt{\left(\frac{{\hat{p}}_1脳{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2脳{\hat{q}}_2}{n_2}\right)}$

For computing the confidence interval, first find the critical value $z_{\frac{伪}{2}}$.

The confidence level is 98% if the level of significance used in the one-railed test is 0.01 (part a.).

Thus, the value of the level of significance for the confidence interval becomes $伪=0.02$.

Hence,

$$\begin{gathered} \frac{伪}{2}=\frac{0.02}{2} \\ =0.01 \end{gathered}$$

The value of $z_{\frac{伪}{2}}$form the standard normal table is equal to 2.33.

Now, the margin of error (E) is equal to:

$$\begin{gathered} E=z_{\frac{伪}{2}}脳\sqrt{\left(\frac{{\hat{p}}_1脳{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2脳{\hat{q}}_2}{n_2}\right)} \\ =2.33脳\sqrt{\left(\frac{0.7733脳0.2267}{150}+\frac{0.1959脳0.8041}{148}\right)} \\ =0.1101 \end{gathered}$$

Substitute the value of E in equation (1) as follows:

$$\begin{gathered} \text{Confidence}鈥�鈥�\text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E鈥�鈥�,鈥�鈥�\left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.7733-0.1959\right)-0.1101鈥�鈥�,鈥�鈥�\left(0.7733-0.1959\right)+0.1101\right) \\ =\left(0.467,0.687\right) \end{gathered}$$

Thus, the 98% confidence interval for the difference between two proportions is.

Since the confidence interval does not include the value of 0 and includes all positive values, it can be concluded that the proportion of patients who are free from headache 15 minutes after oxygen treatment is greater than theproportion of patients who are free from headache 15 minutes after receiving placebo.

c.

From the results, it can be concluded thatthe oxygen treatment is effective because the proportion of patients free from headache after oxygen treatment is greater than the proportion of patients free from headache after receiving a placebo.