An Air Quality instrument logs 0 when standards are not met and 1 when standards are met. The log is saved to data file. First compute the proportion meeting standards as the mean of Air Quality values. Second, at alpha = 0.10 (sensitive, exploratory), test the hypothesis that proportion of time that air quality meets standards is at least 90% a) The pvalue of 0.062 indicates that the data provide weak evidence against Ho: \pi \geq 0.90. HO is rejected at \propto = 0.10 b) The pvalue of 0.022 indicates that the data provide strong evidence against HO: \pi \geq 0.90 HO is rejected at \propto = 0.10. The status quo has changed. c) The pvalue of 0.006 indicates that the data provide overwhelming evidence against HO: \pi \geq 0.90 HO is rejected at \propto = 0.10. Send out an air quality alert d) None of the answers are correct. e) The pvalue of 0.966 indicates that the data provide insignificant evidence against HO: \pi \geq 0.90 HO is not rejected at \propto = 0.10. The status quo remains.
Time Air Quality 0:00:00 1 0:00:30 1 0:01:00 1 0:01:30 1 0:02:00 1 0:02:30 1 0:03:00 1 0:03:30 1 0:04:00 1 0:04:30 1 0:05:00 1 0:05:30 1 0:06:00 1 0:06:30 1 0:07:00 1 0:07:30 1 0:08:00 1 0:08:30 0 0:09:00 0 0:09:30 0 0:10:00 1 0:10:30 1 0:11:00 1 0:11:30 0 0:12:00 1 0:12:30 0 0:13:00 1 0:13:30 0 0:14:00 1 0:14:30 1 0:15:00 0 0:15:30 1 0:16:00 0 0:16:30 1 0:17:00 1 0:17:30 0 0:18:00 1 0:18:30 1 0:19:00 1 0:19:30 1 0:20:00 1 0:20:30 1 0:21:00 1 0:21:30 1 0:22:00 1 0:22:30 1 0:23:00 1 0:23:30 1
Answer:-
Given That:-
First compute the proportion meeting standards as the mean of Air Quality values. Second, at alpha = 0.10 (sensitive, exploratory), test the hypothesis that proportion of time that air quality meets standards is at least 90%
Here
Given
n = 50
x = 49
=0.98
= 1.89
p-value = p(z < 1.89)
= 0.966
(e) The pvalue of 0.966 indicates that the data provide insignificant evidence against is not rejected at = 0.10. The status quo remains unchanged.
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