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A. Identify the test stastics B. Identify the P value C. Test the claim by const

ID: 3323196 • Letter: A

Question

A. Identify the test stastics

B. Identify the P value

C. Test the claim by constructing an appropriate confidence interval.

In a study of treatments for very painful "cluster" headaches, 143 patients were treated with oxygen and 146 other patients were given a placebo consisting of ordinary air. Among the 143 patients in the oxygen treatment group, 111 were free from headaches 15 minutes afteteatment. Among the 146 patients given the placebo, 34 were free from headaches 15 minutes after treatment. Use a 0.05 significance level to test the claim that the oxygen treatment is effective. 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 hypothesis test? O B. Ho P1 SP2 H1 : p1 #p2 0 A. Ho : p1 #p2 H1 P1 P2 OF. Ho : p1=p2 H1 : p1 #p2 H1 P1 P2 D. Ho: p1 = p2 H1 P1P2

Explanation / Answer

Solution:

a) Given that x1 =111, x2 = 34,
n1 = 143, n2 = 146

p1 = 111/143 = 0.776
p2 = 34/146 = 0.233

D. Test Hypothesis:
H0: p1 = p2
H1: p1 > p2

Pooled proportion p = (p1 * n1 + p2 * n2) / (n1 + n2)

= [0.776*143 + 0.233*146] / (143 + 146)

= 0.502

Std Error SE = sqrt[p * ( 1 - p ) * ((1/n1) + (1/n2)) ]

= sqrt[0.502 * ( 1 - 0.502 ) * [ (1/143) + (1/146) ] ]

= 0.059

Test statistic z = (p1 - p2) / SE = (0.776 - 0.233) / 0.059 = 9.203

b) p value for z = 9.203 is 0.000

p value is 0.000 and sig level is 0.01 so reject the Null Hyp . There is enough evidence to support the claim that cure rate with oxygen treatment is higher

98% confidence level

alpha = 1 - 0.95 = 0.05

critical probability = 1 - alpha/2 = 0.975

Critical value for cumulative probability 0.975 is = +1.959

Std Error = 0.059

Margin of Error = Critical value * SE = 1.959 * 0.059 = 0.116

c) Confidence interval = (p1 - p2) ± 0.116 = (0.776 - 0.233) ± 0.116 = (0.427, 0.659)

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