Mathematics, 21.07.2021 21:10, amychan123

A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The population standard deviation is $5.92. Round your answers to two decimal places.

(a) State the point estimate for the population mean cost of fast food bills at this restaurant $

(b) Calculate the 95% margin of error. $

(c) State the 95% confidence interval for the population mean cost of fast food bills at this restaurant.

$
≤ µ ≤ $

(d) What sample size is needed if the error must not exceed $1.00?

n =

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First, we find the point estimate, given by the sample mean. Then, with this, and the standard deviation of the population given, we can find the margin of error, and then, we can find the confidence interval and the minimum sample size necessary.

Doing this, we get that:

a) The point estimate for the population mean cost of fast food bills at this restaurant is $18.21.

b) The 95% margin of error is $1.64.

c) The 95% confidence interval for the population mean cost of fast food bills at this restaurant is: $16.57 ≤ µ ≤ $19.85.

d) The sample size needed is 135.

Question a:

The point estimate for the population mean is the sample mean, which is of $18.21.

The point estimate for the population mean cost of fast food bills at this restaurant is $18.21.

Question b:

We have to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a p-value of , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{5.92}{\sqrt{50}} = 1.64$

The 95% margin of error is $1.64.

The lower end of the interval is the sample mean subtracted by M. So it is 18.21 - 1.64 = 16.57

The upper end of the interval is the sample mean added to M. So it is 18.21 + 1.64 = 19.85

The 95% confidence interval for the population mean cost of fast food bills at this restaurant is: $16.57 ≤ µ ≤ $19.85.

(d) What sample size is needed if the error must not exceed $1.00?

This is n for which M = 1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1 = 1.96\frac{5.92}{\sqrt{n}}$

$\sqrt{n} = 1.96*5.92$

$(\sqrt{n})^2 = (1.96*5.92)^2$

n = 134.6

Rounding up:

The sample size needed is 135.

For a question in which you find a confidence interval using the z-distribution, you can check

To find the minimum sample size for a confidence interval, you can check

answered: Guest

where are the choices?

answered: Guest

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

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