(Please provide a breakdown and describe/show step by step details on how you arrived the answers so I can learn - thanks)

QUESTION: A Sound Quality instrument logs 0 when standards are not met and 1 when standards are met. The log is saved to file DATA (see below 2 columns - Data and Sound Quality).

- First, compute the proportion meeting standards as the mean of Sound Quality values.

- Second, at alpha 0.10 (sensitive, exploratory), test the hypothesis that proportion of times that sound quality meets standards is at least 90%.

(Note: The sample proportion is the mean of the Sound Quality values. Compute the mean to 4 decimal place accuracy. Test (null) H: 0.90.)

A) The p-value of 0.022 indicates that the data provide strong evidence against (null) H: 0.90. (null) H is rejected at (alpha) = 0.10. The status quo has changed.

B) The p-value of 0.966 indicates that the data provide insignificant evidence against H: 0.90. H is not rejected at (alpha) = 0.10. The status quo remains unchanged.

C) The p-value of 0.062 indicates that the data provide weak evidence against H: 0.90. H is rejected at = 0.10.

D) The p-value of 0.006 indicates that the data provide overwhelming evidence

against H: 0.90. H is rejected at = 0.10. Send out a sound quality alert.

E) None of the answers are correct.

DATA: (Time & Sound Quality)

Time Sound 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