An Air Quality instrument logs 0 when standards are not met and 1 when standards are met. The log is in DATA below. 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	0
0:01:30	1
0:02:00	1
0:02:30	1
0:03:00	0
0:03:30	1
0:04:00	1
0:04:30	1
0:05:00	1
0:05:30	0
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	0
0:14:00	0
0:14:30	1
0:15:00	1
0:15:30	1
0:16:00	1
0:16:30	1
0:17:00	1
0:17:30	0
0:18:00	0
0:18:30	0
0:19:00	1
0:19:30	1
0:20:00	1
0:20:30	1
0:21:00	1
0:21:30	0
0:22:00	1
0:22:30	1
0:23:00	0
0:23:30	1

POSSIBLE ANSWER OPTIONS:

1	The pvalue of 0.062 indicates that the data provide weak evidence against H0: 0.90. H0 is rejected at = 0.10.
2	The pvalue of 0.006 indicates that the data provide overwhelming evidence against H0: 0.90. H0 is rejected at = 0.10. Send out an air quality alert.
3	None of the answers match my calculation.
4	The pvalue of 0.966 indicates that the data provide insignificant evidence against H0: 0.90. H0 is not rejected at = 0.10. The status quo remains unchanged.
5	The pvalue of 0.022 indicates that the data provide strong evidence against H0: 0.90. H0 is rejected at = 0.10. The status quo has changed.