Question

In a study of woments for very unter" headaches, 16 patients were treated with oxygen and 146 other patients were given a placebo coordinary Among the patients in the skypen treatment group, 123 were free from headaches 15 minutes after treatment Among the 196 ports on the placebo, 28 were free from headaches 15 minutes after treatment the Oscancerved to test the claim that the wygen att etective Complete para toch below a. Test the coming a hypothesis test Consider the first sample to be the sample of patients treated with oxygen and the second sample to be the sample of patients given a placebo What are there and some hypotheses to the otheister О Р'Р ОА , РР, H₂ Pip OD HP Ос. н PP HP ок гра Hi pipi OL HP Round two decimal places as needed) dently the wa Click to see your answers) 1:57 PM e Type here to search o A 2 9 3 E W Y R Ener

Answer 1

Let the proportion of patients who were free from headache on treatment with oxygen be P1 and that with placebo be P2.

Then,

Null hypothesis (Ho) : P1 = P2

Or, P1-P2 = 0

Alternative hypothesis (H1) : P1 > P2

Or, P1-P2 > 0

Test statistic is given by -

$z = \frac{p_1-p_2}{\sqrt{p*(1-p)*(\frac{1}{n_1}+\frac{1}{n_2}})}$

where, P1 is the sample proportion of patients who recovered after treatment with oxygen = 123/143 = 0.86

P2 is the sample proportion of patients who recovered after treatment with placebo = 28/146 = 0.19

n1 is the sample size of the first group = 143

n2 is the sample size of the second group = 146

p is the Pooled proportion = (123+28)/(143+146) = 0.52

So, the value of the test statistic will be -

$z = \frac{0.86-0.19}{\sqrt{0.52*(1-0.52)*(\frac{1}{143}+\frac{1}{146}})}$

  $= \frac{0.67}{0.059}$

= 11.36

P value = P(z > 11.36)

Area under the z curve on the right side of z = 11.36

= P(z < -11.36) since the z curve is symmetric

= 0.0000...

P(z < -11.36) can be obtained using the formula =NORMSDIST(-11.36) in excel and is nearly zero.

That is P value will be very close to zero.

Since, the P value is less than level of significance (0.05), so, reject the null hypothesis. There is sufficient evidence to support the claim that the cure rate of oxygen is higher than the cure rate for those given placebo.

90% confidence interval for P1-P2 is given by -

$[(p_1-p_2) \pm z_{0.10} * \sqrt{p*(1-p)*(\frac{1}{n_1}+\frac{1}{n_2}})]$

$=[0.67 \pm 1.28 * 0.059]$

$=[0.67 \pm 0.076]$

= [ 0.594, 0.746]

Because the confidence interval limit excludes 0, it appears that the two cure rates are significantly different. Because the confidence interval limit include positive values, it appears that the cure rate is higher for the oxygen treatment than the placebo.

2) Proportion of people having common attribute for the first sample be P1 and for the second sample be P2.

Null hypothesis (Ho) : P1 = P2

Alternative hypothesis (H1) : $P_1 \neq P_2$

Test statistic is given by -

$z = \frac{p_1-p_2}{\sqrt{p*(1-p)*(\frac{1}{n_1}+\frac{1}{n_2}})}$

where, p1 = 15/30 = 0.50

p2 = 1589/2200 = 0.72

p = pooled proportion = (15+1589)/(30+2200) = 0.72

n1 = 30

n2 = 2200

Hence, the value of the test statistic will be -

$= \frac{0.50-0.72}{\sqrt{0.72*(1-0.72)*(\frac{1}{30}+\frac{1}{2200})}}$

$= \frac{-0.22}{0.082}$

= -2.68

Now, the Critical values of z for two tailed test at 0.01 level of significance = (-2.576, 2.576)

(It can be obtained from the z table by finding the z for which area is approximately equal to 0.01/2)

Since, the value of the test statistic (= -2.68) < Critical values of z (= -2.576),

The test statistic is in the critical region,so, reject the null hypothesis. There is sufficient evidence to conclude that $P_1 \neq P_2$

The 99% confidence interval for P1-P2 is given by:

$[(p_1-p_2) \pm z_{0.01/2} * \sqrt{p*(1-p)*(\frac{1}{n_1}+\frac{1}{n_2}})]$

$=[(-0.22) \pm 2.58* 0.082]$

$=[(-0.22) \pm 0.212]$

= [0.008, 0.432]

Since, zero is excluded (or outside) the interval, it indicates to reject the null hypothesis.

The results from the hypothesis test and confidence interval are same, since, the hypothesis test suggest that $P_1 \neq P_2$ and the confidence interval suggest that the $P_1 \neq P_2$ .