A study was performed looking at the effect of mean ozone exposure on change in
ID: 3155250 • Letter: A
Question
A study was performed looking at the effect of mean ozone exposure on change in pulmonary function. Fifty hikers were recruited into the study; 25 study participants hiked on days with low-ozone exposure and 25 hiked on days with high-ozone exposure. The change in pulmonary function after a 4-hour hike was recorded for each participant. The results are given in the following table:
Comparison of change in FEV on high ozone vs. low ozone days
Ozone level
Mean change in FEV
sd
n
High
0.101
0.253
25
Low
0.042
0.106
25
What test can be used to determine whether the mean change in FEV differs between the high ozone and low ozone days?
Implement the test in the problem above and report a p-value (two-tailed).
Determine a 95% CI for: - true mean change in FEV on high ozone days - true mean change in FEV on low ozone days - true mean difference in change in FEV on high ozone days vs. low ozone days
PLEASE SHOW WORK - NO MINITAB! Thank you!
Comparison of change in FEV on high ozone vs. low ozone days
Ozone level
Mean change in FEV
sd
n
High
0.101
0.253
25
Low
0.042
0.106
25
Explanation / Answer
What test can be used to determine whether the mean change in FEV differs between the high ozone and low ozone days?
It is an independent two sample t test.
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Implement the test in the problem above and report a p-value (two-tailed).
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 0.101
X2 = 0.042
Calculating the standard deviations of each group,
s1 = 0.253
s2 = 0.106
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 25
n2 = sample size of group 2 = 25
Thus, df = n1 + n2 - 2 = 48
Also, sD = 0.054861644
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 1.075432589
Also, using p values, as this is two tailed,
p = 0.28755719 [ANSWER, P VALUE]
As P > 0.05, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Hence, there is no significant difference between the mean change in FEV for high and low ozone levels. [CONCLUSION]
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Determine a 95% CI for: - true mean change in FEV on high ozone days - true mean change in FEV on low ozone days - true mean difference in change in FEV on high ozone days vs. low ozone days
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
By table/technology,
t(alpha/2) = 2.010634758
Thus,
lower bound = [X1 - X2] - t(alpha/2) * sD = -0.051306729
upper bound = [X1 - X2] + t(alpha/2) * sD = 0.169306729
Thus, the confidence interval is
( -0.051306729 , 0.169306729 ) [ANSWER]
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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
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