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):
57.4 | 56.3 | 53 | 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 bar = ? %
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 + ? %
a. %
Create the following table.
data | data-mean | (data - mean)2 |
57.4 | -4.86 | 23.6196 |
56.3 | -5.96 | 35.5216 |
53 | -9.26 | 85.7476 |
65.7 | 3.44 | 11.8336 |
59.0 | -3.26 | 10.6276 |
64.7 | 2.44 | 5.9536 |
70.1 | 7.84 | 61.4656 |
64.7 | 2.44 | 5.9536 |
53.5 | -8.76 | 76.7376 |
78.2 | 15.94 | 254.0836 |
Find the sum of numbers in the last column to get.
So standard deviation is %
b. t value for 9 df is TINV(0.1,9)=1.8
So Margin of Error is
Hence CI is
So lower limit is 57.7%
Upper limit is 66.9%
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