Homework SourseTwo popular drugs used for the treatment of depression are R
Two popular drugs used for the treatment of depression are Resithan and Exemor. A random sample of 427 depressed individuals is selected and treated with Resithan, and 151 find relief from their depression. A random sample of 571 depressed individuals is independently selected from the first sample and treated with Exemor, and 248 find relief from their depression. Can we conclude, at the 0.1 level of significance, that the proportion p1of depressed individuals taking Resithan who find relief from depression is less than the proportion p2of all depressed individuals taking Exemor who find relief from depression?
Perform a one-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table
The null hypothesis
H0:
The alternative hypothesis
H1 :
The type of test statistic: Choose One
Z, t, Chi Square, F
The value of the test statistic: (Round to at least three decimal places.)
The p-value: (Round to at least three decimal places.)
Can we conclude that the proportion of depressed individuals taking Resithan who find relief is less than the proportion taking Exemor who find relief?
Yes or No
| The null hypothesis | H0: |
| The alternative hypothesis | H1 : |
| The type of test statistic: Choose One | Z, t, Chi Square, F |
| The value of the test statistic: (Round to at least three decimal places.) | |
| The p-value: (Round to at least three decimal places.) | |
| Can we conclude that the proportion of depressed individuals taking Resithan who find relief is less than the proportion taking Exemor who find relief? | Yes or No |
Solution
The null hypothesis
Ho: p1=p2
The alternative hypothesis:
H1 : p1<p2
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The type of test statistic:
Z
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The value of the test statistic: (Round to at least three decimal places.)
p1=151/427 = 0.35363
p2=248/571 = 0.4343257
Z=(p1-p2)/sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)
=(0.35363-0.4343257)/sqrt(0.35363*(1-0.35363)/427+0.4343257*(1-0.4343257)/571)
=-2.596905
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The p-value: (Round to at least three decimal places.)
It is a left-tailed test.
So the p-value= P(Z<-2.596905) =0.0047 (from standard normal table)
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Can we conclude that the proportion of depressed individuals taking Resithan who find relief is less than the proportion taking Exemor who find relief?
Yes.
