In: Statistics and Probability
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 |
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 ]
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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 ]
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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