Question

What percentage of hospitals provide at least some charity care? Based on a random sample of hospital reports from eastern states, the following information is obtained (units in percentage of hospitals providing at least some charity care):

57.2

56.3

53.4

65.7

59.0

64.7

70.1

64.7

53.5

78.2

Assume that the population of x values has an approximately normal distribution.

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean percentage x and the sample standard deviation s. (Round your answers to one decimal place.)

x =	%
s =	%

(b) Find a 90% confidence interval for the population average μ of the percentage of hospitals providing at least some charity care. (Round your answers to one decimal place.)

lower limit	%
upper limit	%

Answer 1

Part a

Xbar = 62.3%

S = 7.9%

(by using calculator)

Part b

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 62.28

S = 7.932184791

n = 10

df = n – 1 = 9

Confidence level = 90%

Critical t value = 1.8331

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 62.28 ± 1.8331*7.932184791/sqrt(10)

Confidence interval = 62.28 ± 4.5981

Lower limit = 62.28 ± 4.5981 = 57.7

Upper limit = 62.28 ± 4.5981 = 66.9

Confidence interval = (57.7, 66.9)

Lower limit = 57.7%

Upper limit = 66.9%