Question 188223: CAN SOMEONE PLEASE HELP ME IN STATISTICS? THANK YOU SO MUCH!!
Does cymbolta reduce the risk of a heart attack? In a kentucky study, researchers gave cymbolta to 2,450 people and an inactive substitute to 2,370 people (average age 58). After 5 years, 58 of the cymbolta group had suffered a heart attack, compared with 97 for the inactive pill. (a) State the appropriate null and alternative hypotheses. (b) Interpret the results at a = .01 (c) Is normality assured? (d) Is the difference large enough to be important?
Hypothesis test for two independent proportions
p1
0.0237
58/2450
58.
2450
p2
0.0409
97/2370
97
2370
pc
0.0322 p (as a decimal)
155/4820 p (as a fraction)
155. X
4820 n
-0.0173 difference
0. hypothesized difference
0.0051 std. error
-3.39 z
.0003 p-value (one-tailed, lower)
-0.0272 confidence interval 95.% lower
-0.0073 confidence interval 95.% upper
0.01 half-width
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Does cymbolta reduce the risk of a heart attack? In a kentucky study, researchers gave cymbolta to 2,450 people and an inactive substitute to 2,370 people (average age 58). After 5 years, 58 of the cymbolta group had suffered a heart attack, compared with 97 for the inactive pill.
(a) State the appropriate null and alternative hypotheses.
Ho: u(med)-u(nomed) = 0
Ha: u(med)-u(nomed) < 0
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(b) Interpret the results at a = .01
Since the p-level (0.0003) is less than 1%, Reject Ho.
The medicine makes a difference.
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(c) Is normality assured?
I'll leave that to you.
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(d) Is the difference large enough to be important?
Yes since the p-value shows that less than 0.03% of test
results could have given stonger evidence for rejecting Ho.
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cheers,
Stan H>
Hypothesis test for two independent proportions
p1
0.0237
58/2450
58.
2450
p2
0.0409
97/2370
97
2370
pc
0.0322 p (as a decimal)
155/4820 p (as a fraction)
155. X
4820 n
-0.0173 difference
0. hypothesized difference
0.0051 std. error
-3.39 z
.0003 p-value (one-tailed, lower)
-0.0272 confidence interval 95.% lower
-0.0073 confidence interval 95.% upper
0.01 half-width
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