Question

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 33 minutes. From past experience, it is known that the population standard deviation equals 10 minutes. a. Compute the standard error of the mean. b. What can be said about the sampling distribution for the average amount of time customers spent in the restaurant? Be sure to explain your answer. c. With a .95 probability, what statement can be made about the size of the margin of error? d. Construct a 95% confidence interval for the true average amount of time customers spent in the restaurant. e. With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less?

Answer 1

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±$Z_{1-\alpha /2}$*(σ/√n)]

The margin of error of the 95% interval is:

$Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96$

d= $Z_{1-\alpha /2}$*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±$Z_{1-\alpha /2}$*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= $Z_{1-\alpha /2}$*(σ/√n)

$\frac{d}{Z_{1-\alpha /2}}$= σ/√n

√n*($\frac{d}{Z_{1-\alpha /2}}$)= σ

√n= σ* ($\frac{Z_{1-\alpha /2}}{d}$)

n=( σ* ($\frac{Z_{1-\alpha /2}}{d}$))²

n= (10*$\frac{1.96}{2.5}$)²= 61.47≅ 62 customers

I hope this helps!