Question

In: Statistics and Probability

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.

95 172 130 98 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

Solutions

Expert Solution

TRADITIONAL METHOD
given that,
sample mean, x =107.2
standard deviation, s =29.1
sample size, n =9
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 29.1/ sqrt ( 9) )
= 9.7
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 1.86
margin of error = 1.86 * 9.7
= 18
III.
CI = x ± margin of error
confidence interval = [ 107.2 ± 18 ]
= [ 89.2 , 125.2 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =107.2
standard deviation, s =29.1
sample size, n =9
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 1.86
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 107.2 ± t a/2 ( 29.1/ Sqrt ( 9) ]
= [ 107.2-(1.86 * 9.7) , 107.2+(1.86 * 9.7) ]
= [ 89.2 , 125.2 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 89.2 , 125.2 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
Answer:
a.

sample mean =107.2 thousand dollars
sample standard deviation =29.1 thousand dollars

b.
confidence interval =89.2 , 125.2 thousand dollars


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