Question

One method for measuring air pollution is to measure the concentration of carbon monoxide, or CO, in the air. Suppose Nina, an environmental scientist, wishes to estimate the CO concentration in Budapest, Hungary. She randomly selects 48 locations throughout the city measures the CO concentration at each location. Based on her 48 samples, she computes the margin of error for a 95% t-confidence interval for the mean concentration of CO in Budapest, in g/m3, to be 4.28 What would happen to the margin of error if Nina decreases the confidence level to 90%? Nina increases the confidence level to 99%? Nina decreases the sample size to 34 locations? Nina increases the sample size to 70 locations? Answer Bank Decrease Stay the sameIncrease

Answer 1

a) Nina decreases the confidence level to 90%? (Decrease)

b) Nina decreases the sample size to 34 locations? (Increase)

c) Nina increases the sample size to 70 locations? (Decrease)

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n=48 represent the original sample size

Confidence =95% or 0.95

ME=4.28 represent the margin of error.

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

And the margin of error is given by the following expression:

`ME= t_(\alpha/2)(s)/(√(n))` (4)

Based on the formula (4) we can answer all the questions involved:

a) Nina decreases the confidence level to 90%?

On this case the value for
`t_(\alpha/2)` will also decrease so the margin of error would decrease.

b) Nina decreases the sample size to 34 locations?

If we analyze the original sample size of 48 we see that if we reduce the value of n to 34, the margin of error would increase, because n is on the denominator of the margin of error.

c) Nina increases the sample size to 70 locations?

If we analyze the original sample size of 48 we see that if we increase the value of n to 70, the margin of error would decrease, because n is on the denominator of the margin of error.