An environmental chemist is performing a study of iron in atmospheric particulat
ID: 3223495 • Letter: A
Question
An environmental chemist is performing a study of iron in atmospheric particulate measured downwind from a steel mill. She is concerned that the iron readings on a windy day will be lower compared to calm days. She obtained observations on 30 randomly chosen days during the period of peak operation of the mill and compare measurements taken on days when the wind is calm with measurements taken on windy days.
Windy Days
0.25 0.29 0.30 0.43 0.45 0.50 0.60 0.65 0.69 0.74 0.80 0.87 0.87 0.89 0.91 0.92 0.93 0.95 1.01 1.03 1.16
Calm Days
0.68 0.74 0.88 0.89 0.97 1.00 1.17 1.25 1.27
(a) What hypothesis can be tested about the effect of wind on the measurement of iron in atmospheric particulate?
(b) Are you willing to assume that the underlying variances are equal? Justify your answer
Explanation / Answer
Solution:-
For windy days -
0.25,0.29,0.30,0.43,0.45,0.50,0.60,0.65,0.69,0.74,0.80,0.87,0.87,0.89,0.91,0.92,0.93,0.95,1.01,1.03,1.16
Total numbers, n1 = 21
Standard deviation, s1 = 0.2613
Mean, x1 = 0.72571428571429
Calm days
0.68, 0.74, 0.88, 0.89, 0.97, 1.00, 1.17, 1.25, 1.27
Total number, n2 = 9
Standard deviation, s2 = 0.1998
Mean, x2 = 0.98333333
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: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is let say0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(.26132/21) + (0.19982/9)] = 0.08767
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (.26132/21 + 0.19982/9)2 / { [ (.26132 / 21)2 / (20) ] + [ 0.19982 / 9)2 / (8) ] }
DF = 0.000059 /(0.00000005285 + 0.000002459)
= 0.000059 / 0.00000251185 = 23.4886637339
t = [ (x1 - x2) - d ] / SE = [ (0.72571428571429 - 0.98333333) - 0 ] / 0.08767 = -2.9385085466
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic
We use the t Distribution Calculator
The P-Value is .007426.
The result is significant at p < .10.
Interpret results. Since the P-value (0.007426) is less than the significance level (0.10), we cannot accept the null hypothesis.