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Vaselesa [24]
3 years ago
10

One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was rev

ealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99% confidence interval for the population mean hours spent watching television per month.
Mathematics
2 answers:
dmitriy555 [2]3 years ago
8 0

Answer:

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.575*\frac{32}{\sqrt{108}} = 7.93

The lower end of the interval is the mean subtracted by M. So it is 151 - 7.93 = 143.07 hours

The upper end of the interval is the mean added to M. So it is 151 + 7.93 = 158.93 hours

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

PtichkaEL [24]3 years ago
4 0

Answer:

try

Step-by-step explanation:

just try people

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