# Lisa's science fair project was to estimate the mean amount of chemicals in her city's water...

## Question:

Lisa's science fair project was to estimate the mean amount of chemicals in her city's water supply. At first, she had decided to use a random sample of 15 observations. But her teacher asked her to take 35 observations. The mean and standard deviation from 35 observations turned out to be approximately the same as those from 15 observations.

Is there any advantage in using 35 observations instead of 15 observations?

A. There is no advantage. Because the mean and the standard deviation are about the same, the confidence interval computed using 35 observations should be approximately the same as that computed using 15 observations.

B. There is no advantage. In fact, the 20 extra observations will increase the likelihood of error.

C. There is no advantage. Because she took 35 observations instead of 15 observations, the confidence interval using 35 observations will be wider than that using 15 observations.

D. There is some advantage. Because she took 35 observations instead of 15 observations, the confidence interval using 35 observations will be narrower than that using 15 observations.

E. There is some advantage. With 35 observations, she will be able to compute an exact z-confidence interval instead of an approximate t-confidence interval.

## Confidence Interval

The confidence interval is used to estimate the population mean. The interval contains the point estimate, which is the sample mean, and the marign of error for the estimate.

## Answer and Explanation:

The answer is D and E. There is some advantage. Because she took 35 observations instead of 15 observations, the confidence interval using 35 observations will be narrower than that using 15 observations, and she will be able to use the Z-score which is smaller for the same confidence level that the t-value.

The confidence interval is the point estimate, taken from the sample mean, and the margin of error which is the Z-score times the standard error of the mean. The equation for margin of error is:

{eq}E = Z \dfrac{s}{\sqrt{n}} {/eq}

Where:

• {eq}E {/eq} is the margin of error
• {eq}Z {/eq} is the Z-score which is determined by the confidence interval.
• {eq}s {/eq} is the sample standard deviation.
• {eq}n {/eq} is the sample size.

Since sample size is in the denominator of the equation for the margin of error, and twice the margin of error is the interval width, the interval width for a sample size of {eq}35 {/eq} will be smaller than the interval width of {eq}15 {/eq}.

Using a t-lookup table for a sample size of {eq}15 {/eq} and {eq}95\% {/eq} confidence level, the t-value is {eq}2.14 {/eq}. For a Z-score of {eq}95\% {/eq} confidence, from a Z lookup table the Z-score is {eq}1.96 {/eq}. Therefore, by increasing the sample size to {eq}35 {/eq}, where the Z-score replaces the t-value in the equation for margin of error, the confidence interval is narrower.

#### Learn more about this topic:

Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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