SOLUTION: Consider a study conducted to evaluate the effectiveness of a new treatment to stop smoking. Before the study, it was known the average time to quit was 2.3 months with a standard
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Question 1198099: Consider a study conducted to evaluate the effectiveness of a new treatment to stop smoking. Before the study, it was known the average time to quit was 2.3 months with a standard deviation of 1.1. Using a sample of 50 adults on the new treatment, the sample average time was 1.9 months. Questions: 6) what is the test value for the sample of 50 adults? Question 7: The p-value for the test value should be? Question 8: which of the following statements about the significance this study at a 0.05 level is correct?: A) The p-valueof test value is greater than .05, it is significant. B) The p-value of the test value is greater than, 0.05, it is not significant. C) The p-value of the test value is less than .05 it is significant. D) The p-value of the test value is less than .05, it is not significant
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Normal Distribution: µ =2.3mo and σ = 1.1
Sample of 50: x̄ = 1.9mo
6) test statistic = z = = -2.5713
7) p(-2.5713) = normacdf(-9999, -2.5713, 1, 0) = .0051
8) which of the following statements about the significance of this study
at a 0.05 level is correct?
C) The p-value of the test value is less than .05 it is significant.
Significant, in that, the Evidence indicates the claim µ =2.3mo be rejected.
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