Question

You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 12 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
19 12 7 8 16 12 21 26 28 19 10 12
a. To compute the confidence interval use a
?
distribution.

b. With 95% confidence the population mean number of visits per physical therapy patient is between
and
visits.

c. If many groups of 12 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About
percent of these confidence intervals will contain the true population mean number of visits per patient and about
percent will not contain the true population mean number of visits per patient.

Answer 1

In order to calculate the confidence Interval, what we need is the t distribution. The population mean number of visits is 11.667 and 19.9928 visits. 95 percent contains true mean while the other 5 does not.

How to solve for the statistics

We have n = 12

df = n- 1 = 11

a. To compute the confidence interval use a t distribution.

The mean = fx/n

= 15.83

level of significance = 0.05

standard deviation = 6.555

t α/2, df = 2.20

standard error  = 6.555 / √ 12 = 1.8922

margin of error , = 2.2 * 1.8922 = 4.1628

barx - E = 15.83 - 4.1628 = 11.6672

barx + E = 15.83 + 4.1628 = 19.9928

With 95% confidence the population mean number of visits  is between 11.667 and 19.9928 visits.

c. About 95 percent of these confidence intervals will contain true mean.

about __5__percent will not contain the true population

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