In a study of treatments for very painful \"cluster\" headaches, 152 patients we

Question

In a study of treatments for very painful "cluster" headaches, 152 patients were treated with oxygen and 145 other patients were given a placebo consisting of ordinary air. Among the 152 patients in the oxygen treatment group, 117 were free from headaches 15 minutes after treatment. Among the 145 patients given the placebo, 23 were free from headaches 15 minutes after treatment. Use a 0.01 significance level to test the claim that the oxygen treatment is effective. Complete parts (a) through (c) below.

1) Pvalue is (less/than).......so(reject,fail to reject).....there(is sufficient/is not sufficient)

2) because......limits(include/do not include)......rates are (not equal/ equal).....include(only positive/positive and negative/only negative)..... cure rate is ( the same/higher/lower)

Please aswer complete

cluster headaches, 152 patients were treated with oxygen and 145 other patients were given a placebo consisting of ordinary air. Among the 152 patients in the axygen treatment group. 117 oxygen treatment is were free from headaches 15 minutes after treatment Among the 145 patients given the ients given the placebo, 23 were free from headaches 15 minutes after treatment Use a 0 01 significance level to test the claim that the eftective Complete parts (a) through (c) below a. Test the claim using 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 the null and alternative hypotheses for the hvoothesin te

Explanation / Answer

as the treatment should reduce the headache , hence p1 > p2 as the alternative hypothesis , so the correct choice is
For part A the correct answer is E

Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
p = (p1 * n1 + p2 * n2) / (n1 + n2)

where p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.

putting the values

p1 = 117/152 = 0.769
p2 = 23/145 = 0.158

p = (p1 * n1 + p2 * n2) / (n1 + n2)
= (0.769*152 + 0.158*)/(152+145) = 0.45

Compute the standard error (SE) of the sampling distribution difference between two proportions.
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.

SE = sqrt( 0.45 * ( 1 - 0.45 ) * [ (1/152) + (1/145) ] )
= 0.057

The test statistic is a z-score (z) defined by the following equation.
z = (p1 - p2) / SE

(0.769-0.158)/0.057
= 10.71 (z stat) this is the z value

we check the p value from the z table as
P ( Z>10.71 )=1P ( Z<10.71 )=11=0 (p value) this is the p value

as the p value is less than alpha = 0.05 , Hence the results are signficant and we can say that the oxygen treatment is effective

C's answer is C

The standard error of proportion for p1 is

sqrt(p*(1-p)/n) = sqrt(0.769*(1-0.769)/152) =0.034

The standard error of proportion for p2 is

sqrt(p*(1-p)/n) = sqrt(0.158*(1-0.158)/145) = 0.030

Standard Error for Difference = sqrt(0.034^2 + 0.03^2) = 0.045

now the CI for proprtion is

Difference Between the Sample Proportions±z(Standard Error for Difference) , here z = 2.33 , from the z table

(0.769 -0.158) +- 2.33*0.045 , solving for the plus and minus sign the CI is

0.50 , 0.71

1) Pvalue is (less/than)0.05 so(reject).....there(is sufficient)

2) because..0.50 , 0.71..limits(do not include)....zero..rates are (not equal)..CI 0.50 , 0.71...include(only positive)..... cure rate is ( the higher)

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