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2. Collette is self-employed, selling items at Home Interior
parties. She wants to estimate the average amount a client
spends at each party. A random sample of 35 clients \'
receipts gave a sample mean of \$34.70 and historically we
know the population standard deviation of spending at these
parties to be \$4.85. a.Explain why you may assume your data
is normally distributed. b.Find a 90% confidence interval for
the average amount expected to be spent by a given client at
a party. c.Write a brief explanation of the meaning of the
confidence interval in the context of this problem d. For a
party with 35 clients, use part (b) to estimate a range of
dollar values for Collette\'s total sales at that party.
Solution
a.Explain why you may assume your data is normally
distributed.
The sample size is large enough, n = 35, and we know the
population standard deviation, so central limit theorem
would lead us that the sample means are normally
distributed.
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b.Find a 90% confidence interval for the average amount
expected to be spent by a given client at a party.
Note that
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Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 34.7
z(alpha/2) = critical z for the confidence interval =
1.644853627
s = sample standard deviation = 4.85
n = sample size = 35
Thus,
Margin of Error E = 1.34845039
Lower bound = 33.35154961
Upper bound = 36.04845039
Thus, the confidence interval is
( 33.35154961 , 36.04845039 ) [ANSWER]
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c.Write a brief explanation of the meaning of the confidence
interval in the context of this problem
Hence, we are 90% confident that the true mean amount a
client spends at each party is between \$33.35154961 and
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d. For a party with 35 clients, use part (b) to estimate a
range of dollar values for Collette\'s total sales at that party.
We then multiply this confidence interval by 35, so we have
( 33.35154961*35 , 36.04845039*35 )