Question

A physician with a practice is currently serving 300 patients. The physician would like to administer a survey to his patients to measure their satisfaction level with his practice. A random sample of 22 patients had an average satisfaction score of 8.3 on a scale of​ 1-10. The sample standard deviation was 1.9. Complete parts a and b below.

a. Construct a​ 99% confidence interval to estimate the average satisfaction score for the​ physician's practice.

The​ 99% confidence interval to estimate the average satisfaction score is________, ___________ ​(Round to two decimal places as​ needed.)

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A university would like to estimate the proportion of fans who purchase concessions at the first basketball game of the season. The basketball facility has a capacity of 3 comma 500 and is routinely sold out. It was discovered that a total of 210 fans out of a random sample of 500 purchased concessions during the game. Construct a​ 95% confidence interval to estimate the proportion of fans who purchased concessions during the game. The​ 95% confidence interval to estimate the proportion of fans who purchased concessions during the game is ___,____.

Answer 1

Solution :

a ) Given that,

= 8.3

s = 1.9

n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,21 =2.831

Margin of error = E = t_/2,df * (s /n)

= 2.831 * (1.9 / 22)

= 1.15

Margin of error = 1.15

The 99% confidence interval estimate of the population mean is,

- E < < + E

8.3 - 1.15 < < 8.3 + 1.15

7.15 < < 9.45

(7.15, 9.45 )

b) Given that,

n = 500

x = 210

= x / n = 210 / 500= 0.42

1 - = 1 - 0.42 = 0.58

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960 * (((0.42 * 0.58 ) / 500)

= 0.067

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.420 - 0.067 < p < 0.420 + 0.067

0.353 < p < 0.487