SOLUTION: In a program designed to help patients stop smoking, 156 patients were given sustained care, and 84.1% of them were no longer smoking after one month. Use a 0.01 significa
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Question 1156754: In a program designed to help patients stop smoking, 156 patients were given sustained care, and 84.1% of them were no longer smoking after one month. Use a 0.01 significance level to test the claim that 80% of patients stop smoking when given sustained care. I have solved the test statistic and it is correct but I just need the p- value for the hypothesis test. Thanks in advance!
Upper H 0: p equals 0.8
Upper H 1: p not equals 0.8
The test statistic is 1.28.
The p value for this hypothesis is ?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
z = (phat - p)/(sqrt(p*(1-p)/n))
z = (0.841 - 0.8)/(sqrt(0.8*(1-0.8)/156))
z = 1.28022458967168
z = 1.28 is the test statistic
Nice job on getting the correct test statistic
Now use a Z table found in the back of your stats textbook to find the area under the standard normal distribution.
If you do not have a book with you, then you can use a Z table such as this one
https://www.ztable.net/
To find that
P(Z < 1.28) = 0.89973
note how I've highlighted the proper row and column which represent 1.20 and 0.08 combining to 1.28
So,
P(Z > 1.28) = 1-P(Z < 1.28)
P(Z > 1.28) = 1-0.89973
P(Z > 1.28) = 0.10027 approximately
Due to the "not equal" sign in the alternative hypothesis, we have a two tailed test. This means we will double the result to get the area in both tails.
2*0.10027 = 0.20054
The p value is approximately 0.20054
Extra Info:
At the significance level alpha = 0.01, we fail to reject the null. We can only reject the null if the p value is smaller than alpha. Failing to reject the null means we effectively "accept" the null hypothesis (until other data comes along to invalidate it). In other words, we conclude that p = 0.8 is the case. In the context of the problem, we conclude that 80% of patients quit smoking after one month.
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