Time Air Quality 0:DO:00 0:00:30 0 01:00 0.01:30 0:02:00 0:02:30 0:03:00 0:03:30 0:04:00 0:04:30 0:05:00 0:05:30 0:06:00 0:06:30 0:07:00 0:07:30 0:08:00 0:08:30 0:09:00 0:09:30 0:10:00 0:10:30 0:11:00 0:11:30 0:12:00 0:12:30 0:13:00 0:13:30 0:14:00 0:14:30 0:15:00 0:15:30 0:16:00 0:16:30 0:17:00 0:17:30 0:18:00 0:18:30 0:19:00 0:19:30 0:20:00 0:20:30 0:21:00 0:21:30 0:22:00 0:22:30 0:23:00 0:23:30An Air Quality instrument logs 0 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%. O The pvalue of 0.062 indicates that the data provide weak evidence against H0: 11 2 0.90. H0 is rejected at or = 0.10. 0 None of the answers are correct. 0 The pvalue of 0.966 indicates that the data provide insignificant evidence against H0: 11 2 0.90. H0 is not rejected at or = 0.10. The status quo remains unchanged. 0 The pvalue of 0.022 indicates that the data provide strong evidence against H0: 1': 2 0.90. H0 is rejected at O( = 0.10. The status quo has changed. 0 The pvalue of 0.006 indicates that the data provide overwhelming evidence against H0: at 2 0.90. H0 is rejected at o: = 0.10. Send out an air quality alert. v Hide hint for Question 11 The sample proportion is the mean of the Air Quality values. Compute the mean to 4 decimal place accuracy. Test H0: 11 2 0.90