Question

# Do you want to own your own candy store? Wow! With some interest in running your...

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

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 = t0.05,8 =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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