An Air Quality instrument logs O when standards are not met and 1 when standards are met. The log is saved to file DATA. 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 times that air quality meets standards is at least 90%.

DATA:

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 1

0:09:00 1

0:09:30 1

0:10:00 1

0:10:30 1

0:11:00 1

0:11:30 1

0:12:00 1

0:12:30 1

0:13:00 1

0:13:30 1

0:14:00 1

0:14:30 1

0:15:00 1

0:15:30 1

0:16:00 1

0:16:30 0

0:17:00 1

0:17:30 1

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

1-The pvalue of 0.022 indicates that the data provide strong evidence against HO:

TT0.90. HO is rejected at ex = 0.10. The status quo has changed.

2-The pvalue of 0.062 indicates that the data provide weak evidence against HO: TT

0.90. HO is rejected at ex= 0.10.

3- The pvalue of 0.966 indicates that the data provide insignificant evidence against HO: n: 0.90. HO is not rejected at ex= 0.10. The status quo remains unchanged.

4- The pvalue of 0.006 indicates that the data provide overwhelming evidence against HO: n: 0.90. HO is rejected at ex= 0.10. Send out an air quality alert.

5-None of the answers are correct.