SOLUTION: The A survey of 41 randomly selected "iPhone" owners showed that the purchase price has a mean of $414 with a sample standard deviation of $180. a. Compute the standard er

Question 1193207: The A survey of 41 randomly selected "iPhone" owners showed that the purchase price has a mean of $414 with a sample standard deviation of $180.

a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.)

b. Compute the 80% confidence interval for the mean. (Round the final answers to 2 decimal places.)

The confidence interval is between $
and $

c. How large a sample is needed to estimate the population mean within $13 at a 80% degree of confidence? (Round the final answer to the nearest whole number.)

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Part (a)

n = sample size = 41
s = sample standard deviation = 180

SE = standard error
SE = s/sqrt(n)
SE = 180/sqrt(41)
SE = 28.111277139949 approximately
SE = 28

Answer: 28

==============================================================

Part (b)

Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
which should be similar to what's found in the appendix section of your stats textbook.
It's labeled a T table, but we'll be using a particular Z value in the bottom row highlighted in blue.

We can use Z here because n = 41 makes n > 30 true.

In the bottom row highlighted in blue, we have z = 1.282 when we're dealing with an 80% confidence level.

E = margin of error
E = z*(standard error)
E = z*SE
E = 1.282*28.111277139949
E = 36.0386572934147
E = 36.038657
which is approximate

L = lower bound of confidence interval
L = xbar - E
L = 414 - 36.038657
L = 377.961343
L = 377.96

U = upper bound of confidence interval
U = xbar + E
U = 414 + 36.038657
U = 450.038657
U = 450.04

We round to two decimal places because we're dealing with prices, and it's a standard practice to do so anyway with confidence intervals in general.
Also of course, it's due to the instructions stating "Round the final answers to 2 decimal places."

xbar = 414 is the point estimate of the population mean mu.
xbar is the center of the confidence interval

The confidence interval in the format of (L, U) is (377.96, 450.04)
This is equivalent to saying 377.96 < mu < 450.04
We are 80% confident the true population mean price is somewhere between $377.96 and $450.04

Answer: The confidence interval is between $377.96 and $450.04

==============================================================

Part (c)

Let's solve for n in the margin of error formula so we can determine the minimum sample size.
You can skip to the next paragraph if you already have the min sample size formula handy.
E = z*s/sqrt(n)
E*sqrt(n) = z*s
sqrt(n) = z*s/E
n = (z*s/E)^2

We desire the margin of error (E) to be $13 or smaller
Let's say E = 13

We have the same value of s and z as from before.

The min sample size would be:
n = (z*s/E)^2
n = (1.282*180/13)^2
n = 315.0898
n = 316

Notice I rounded UP to the nearest integer even though 315.0898 is much closer to 315 than it is to 316

We can see that when n = 315 we have this error level
E = z*s/sqrt(n)
E = 1.282*180/sqrt(315)
E = 13.001853
The error is slightly over the target of 13, which is what we don't want.

Now let's try n = 316
E = z*s/sqrt(n)
E = 1.282*180/sqrt(316)
E = 12.981264
The error is now either 13 or smaller as we stated at the top of this section.

The last two paragraphs help illustrate why we always round up to the nearest integer when it comes to min sample size problems.

Answer: 316

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