An 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%.

Data: Time Air Quality 0:00:00 1 0:00:30 0 0:01:00 1 0:01:30 1 0:02:00 0 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 0 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 0 0:16:00 0 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 0 0:22:00 1 0:22:30 0 0:23:00 1 0:23:30 1

Here are the choices of my answers:

1) None of the answers are correct.

2) The p value of .062 indicates that the data provide weak evidence against H0: TT is greater than or equal to 0.90. H0 is rejected at x=0.10.

3) The p value of 0.022 indicates that the data provide strong evidence against H0: TT is greater than or equal to 0.90. H0 is rejected at x=0.10. The status quo has changed.

4) The p value of 0.006 indicates that the data provide overwhelming evidence against H0: TT is greater than or equal to 0.90. H0 is rejected at x=0.10. Send out an air quality expert.

5) The p value of 0.966 indicates that the data provide insignificant evidence against H0: TT is greater than or equal to 0.90. H0 is not rejected at x=0.10. The status quo remains unchanged.

I got a z value of -1.6449 by calculator and a z value of -19.95 with Excel. Please help.