Question 1206824: A survey of 45 randomly selected iPhone owners showed that the purchase price has a mean of $426 with a sample standard deviation of $190.
a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.)
Standard error $
28
b. Compute the 95% confidence interval for the mean. (Round the final answers to 2 decimal places.)
The confidence interval is between $
345.02
and $
409.08
c. How large a sample is needed to estimate the population mean within $6 at a 95% degree of confidence? (Round the final answer to the nearest whole number.)
Sample size
3852
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
n = 45 = sample size
s = 190 = sample standard deviation
SE = standard error
SE = s/sqrt(n)
SE = 190/sqrt(45)
SE = 28.323528 approximately
SE = 28 approximately
You have the correct answer. Nice work.
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Part (b)
sigma = population standard deviation = unknown
Despite not knowing sigma, we can use the Z distribution since n > 30 is the case.
For cases n > 30, the T distribution is approximately close to the Z distribution.
At 95% confidence, the z critical value is roughly 1.96
This is something to memorize or look up on a reference sheet/table.
E = margin of error
E = z*( s/sqrt(n) )
E = z*SE
E = 1.96*28.323528
E = 55.514115 approximately
L = lower boundary
L = xbar - E
L = 426 - 55.514115
L = 370.485885
L = 370.49
I'm not sure how you got 345.02
Please post your steps or thought process.
U = upper boundary
U = xbar + E
U = 426 + 55.514115
U = 481.514115
U = 481.51
I'm not sure how you got 409.08
Please post your steps or thought process.
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Part (c)
Let's rearrange the margin of error equation to solve for variable n.
E = margin of error
E = z*s/sqrt(n)
E*sqrt(n) = z*s
sqrt(n) = z*s/E
n = (z*s/E)^2
Plug in the known info, and the desired error E = 6, to get
n = (z*s/E)^2
n = (1.96*190/6)^2
n = 3852.271111111
n = 3853
Notice how I rounded UP to the nearest whole number. This is despite 3852.271111111 being closer to 3852 than 3853.
Why round up? Well let's see what happens when we try n = 3852
E = z*s/sqrt(n)
E = 1.96*190/sqrt(3852)
E = 6.000211 approximately
The error is slightly over 6 when we want it either exactly 6 or smaller.
This shows that n = 3852 is not big enough.
Now try n = 3853
E = z*s/sqrt(n)
E = 1.96*190/sqrt(3853)
E = 5.999432 approximately
We're now slightly under the error threshold we want.
This confirms that n = 3853 works. As n gets bigger, E will get smaller.
Summary:
Your answer is close but it should be 3853
Your professor should replace "round to the nearest whole number" with "round UP to the nearest whole number".
I've seen many students make this mistake.
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