The number of years it took a random sample of 33 former smokers to quit permane
ID: 3237515 • Letter: T
Question
The number of years it took a random sample of 33 former smokers to quit permanently it listed. We know that the population standard deviation is 1.5 years. A health agency claims that the mean time it takes smokers to quit smoking permanently is 6 years. You think it’s more than that. At a=5%, is there enough evidence to reject the health agency’s claim? At the 1% level? Can you test the claim using an appropriate CI? Explain. If your answer is yes, test the claim that way, too.
Smokers
Years to quit smoking
1
10.7
2
3.2
3
12.6
4
9
5
8.7
6
9.2
7
0.7
8
3
9
2.3
10
8.2
11
12.8
12
9.8
13
3.3
14
1.8
15
0.8
16
1
17
2.5
18
8.1
19
20.7
20
12.5
21
9.9
22
2.9
23
7.1
24
8.3
25
5.2
26
7
27
7.8
28
0.1
29
3.5
30
10.5
31
4.2
32
1.1
33
6.1
Smokers
Years to quit smoking
1
10.7
2
3.2
3
12.6
4
9
5
8.7
6
9.2
7
0.7
8
3
9
2.3
10
8.2
11
12.8
12
9.8
13
3.3
14
1.8
15
0.8
16
1
17
2.5
18
8.1
19
20.7
20
12.5
21
9.9
22
2.9
23
7.1
24
8.3
25
5.2
26
7
27
7.8
28
0.1
29
3.5
30
10.5
31
4.2
32
1.1
33
6.1
Explanation / Answer
Soltion:-
The number of years it took a random sample of 33 former smokers (n = 33) to quit permanently it listed. We know that the population standard deviation is 1.5 years (s = 1.5). A health agency claims that the mean time it takes smokers to quit smoking permanently is 6 years( = 6). You think it’s more than that. At a=5% (0.05), is there enough evidence to reject the health agency’s claim? At the 1% (0.01) level?
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 6
Alternative hypothesis: 6
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 1.5 / sqrt(33) = 0.261
DF = n - 1 = 33 - 1 = 32
t = (x - ) / SE = (6.5 - 6)/0.261 = 1.92
where x = sum of the given data divided by 33. that is the sample mean.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 32 degrees of freedom is less than -1.92 or greater than 1.92.
We use the t Distribution Calculator to find P(t < 1.92) = 0.04,
The P-Value is 0.06381.
The result is not significant at p < 0.05.
Interpret results. Since the P-value (0.06381) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Even if we test for significance level of 0.01, still we can not reject the null hypothesis.
Conclusion. Fail o reject the null hypothsis. We have sufficient evidence to prove that the claim that "the mean time it takes smokers to quit smoking permanently is 6 years" is true.