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.
(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.)

(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 
Solution:
x  x^{2} 
93  8649 
177  31329 
132  17424 
92  8464 
75  5625 
94  8836 
116  13456 
100  10000 
85  7225 
x=964  x^{2}=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 =(
x^{2} )  ((
x)^{2} / n ) n 1
=111008(964)^{2}9
/8
=111008103255.1111/8
=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
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