College Statistics

Question

The mayor is interested in finding a 90% confidence interval for the mean number of pounds of trash per person per week that is generated in city. The study included 169 residents whose mean number of pounds of trash generated per person per week was 35.8 pounds and the standard deviation was 6.1 pounds.

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean number of pounds per person per week is between  and pounds. 

c. If many groups of 169 randomly selected members are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about  percent will not contain the true population mean number of pounds of trash generated per person per week. 

Answer 1

Let's begin by defining what we know:

Residents in our sample = n = 169

Mean amount of trash produced per person per week = x-bar = M = 35.8

Standard deviation of our sample = s = 6.1

A. We need to think about the guidelines for selecting the Student's t distribution or the standard normal z distribution.

1. It is advisable to select the Student's t distribution if we do not know the population variance.

2. However, as the number of observations in our sample increases, the Student's t distribution more closely approximates the standard normal z distribution.

3. We have a sample greater than 120. Therefore, I would argue that it does not substantively matter if we use a student's t distribution with degrees of freedom (n-1) 168, or the standard normal z distribution.

4. Given the information above, generate an argument for your selection of distribution.

B. Find a 90% confidence interval.

1. It is intuitive to remember that the confidence interval will be symmetrical on either side of our mean.

2. As such, we will take our mean (35.8), add a value to it, and then subtract a value from the mean, to find our interval range.

3. The formula to find the confidence interval for a population mean is as follows:

Lower bound = sample_mean - (confidence level)*s/sqrt(n)

Upper bound = sample_mean + (confidence level)*s/sqrt(n)

Let's find the values to fill out the formula:

Sample_mean = 35.8

Confidence level = if we we want a 90% confidence interval, we need capture .90 of the distribution within the interval, leaving .05 of the distribution in each tail. The z score for .05 in each tail is 1.645. (if we used a student't t distribution, our value would be 1.654)

Sample standard deviation = s = 6.1

Observations in sample = n = 169

Lower bound = 35.8 - (1.654)*(6.1/sqrt(169))

Upper bound = 35.8 + (1.654)*(6.1/sqrt(169))

C. What percent of future samples (n = 169) would contain the true population mean?

Here we want to think about the idea of repeated samples. Let's pull a quote from "Statistics for Business and Economics" - Newbold, Carlson, and Thorne 8th edition.

"If the population is repeatedly sampled and intervals are calculated... then in the long run [90%] of the intervals would contain the true value of the [population mean]." p. 271

I hope this is helpful!