Question

Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below. Assume that the population of x values has an approximately normal distribution.

Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below. Assume that the population of x values has an approximately normal distribution.

93

177

132

92

75

94

116

100

85

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

x =	thousand dollars
s =	thousand dollars

(b) Find a 90% confidence interval for the population average startup costs μ for candy store franchises. (Round your answers to one decimal place.)

lower limit	thousand dollars
upper limit	thousand dollars

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

x =	thousand dollars
s =	thousand dollars

(b) Find a 90% confidence interval for the population average startup costs μ for candy store franchises. (Round your answers to one decimal place.)

lower limit	thousand dollars
upper limit	thousand dollars

Answer 1

Solution:

x	x2
93	8649
177	31329
132	17424
92	8464
75	5625
94	8836
116	13456
100	10000
85	7225
x=964	x2=111008

a ) The sample mean is

Mean = (x / n) )

=93+177+132+92+75+94+116+100+85 /9

=964/9

=107.1111

Mean = 107.1

The sample standard is S

  S =( x² ) - (( x)² / n ) n -1

=111008-(964)²9 /8

=111008-103255.1111/8

=7752.8889/8

=969.1111

=31.1305

Sample Standard deviation S = 31.1

b ) Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,8 =1.859

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

= 1.859 * (31.1 / 9)

= 19.3

Margin of error = 19.3

The 90% confidence interval estimate of the population mean is,

- E < < + E

107.1 - 19.3 < < 107.1 + 19.3

87.8 < < 126.4

lower limit = 87.8

upper limit =126.4