At the city museum, child admission is $5.40 and adult admission is $8.70 . On Friday, 170 tickets were sold for a total sales o
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Answer:
We conclude that 80% of patients stop smoking when given sustained care.
Step-by-step explanation:
We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.
Let p = <u><em>percentage of patients stop smoking when given sustained care.</em></u>
So, Null Hypothesis, : p = 80% {means that 80% of patients stop smoking when given sustained care}
Alternate Hypothesis, : p
80% {means that different from 80% of patients stop smoking when given sustained care}
The test statistics that would be used here <u>One-sample z test for proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of patients who stop smoking when given sustained care = 80.2%
n = sample of patients = 192
So, <u><em>the test statistics</em></u> =
= 0.07
The value of z test statistics is 0.07.
<u></u>
<u>Also, P-value of the test statistics is given by the following formula;</u>
P-value = P(Z > 0.07) = 1 - P(Z 0.07)
= 1 - 0.52790 = 0.4721
<u>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>
Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that 80% of patients stop smoking when given sustained care.