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Chapter 8: Problem 8

Refer to Exercise \(8.7 .\) What effect does increasing the sample size have on the margin of error?

Short Answer

Expert verified
Answer: Increasing the sample size has the effect of decreasing the margin of error. As the sample size gets larger, the margin of error becomes smaller, meaning that our estimate becomes more precise.

Step by step solution

01

Recall the Margin of Error Formula

The margin of error (ME) is calculated using the formula: ME = Z * (蟽 / 鈭歯) where Z is the Z-score corresponding to a certain confidence level, 蟽 is the population standard deviation, and n is the sample size.
02

Examine the Relationship between Sample Size and Margin of Error

In the formula, ME = Z * (蟽 / 鈭歯), the only variable we are manipulating is the sample size (n). As we increase the value of n, the denominator (i.e., the square root of n) will also increase, thus making the fraction (蟽 / 鈭歯) smaller.
03

Evaluate the Impact of Increasing Sample Size on Margin of Error

When the fraction (蟽 / 鈭歯) becomes smaller, the overall value of the margin of error (ME) will also decrease, because it is equal to Z * (蟽 / 鈭歯). This indicates that increasing the sample size will result in a smaller margin of error.
04

Conclusion

In conclusion, increasing the sample size has the effect of decreasing the margin of error. As the sample size gets larger, the margin of error becomes smaller, meaning that our estimate becomes more precise.

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Most popular questions from this chapter

Independent random samples of \(n_{1}=50\) and \(n_{2}=60\) observations were selected from populations 1 and \(2,\) respectively. The sample sizes and computed sample statistics are given in the table: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 5 & 60 \\\\\text { Sample Mean } & 100.4 & 96.2 \\\\\text { Sample Standard Deviation } & 0.8 & 1.3\end{array}$$ Find a \(90 \%\) confidence interval for the difference in population means and interpret the interval.

Independent random samples of \(n_{1}=40\) and \(n_{2}=80\) observations were selected from binomial populations 1 and 2 , respectively. The number of successes in the two samples were \(x_{1}=17\) and \(x_{2}=23 .\) Find a \(99 \%\) confidence interval for the difference between the two binomial population proportions. Interpret this interval.

In a report of why e-shoppers abandon their online sales transactions, Alison Stein Wellner \(^{8}\) found that "pages took too long to load" and "site was so confusing that I couldn't find the product" were the two complaints heard most often. Based on customers' responses, the average time to complete an online order form will take 4.5 minutes. Suppose that \(n=50\) customers responded and that the standard deviation of the time to complete an online order is 2.7 minutes. a. Do you think that \(x\), the time to complete the online order form, has a mound-shaped distribution? If not, what shape would you expect? b. If the distribution of the completion times is not normal, you can still use the standard normal distribution to construct a confidence interval for \(\mu\), the mean completion time for online shoppers. Why? c. Construct a \(95 \%\) confidence interval for \(\mu,\) the mean completion time for online orders.

In a study to compare the effects of two pain relievers it was found that of \(n_{1}=200\) randomly selectd individuals instructed to use the first pain reliever, \(93 \%\) indicated that it relieved their pain. Of \(n_{2}=450\) randomly selected individuals instructed to use the second pain reliever, \(96 \%\) indicated that it relieved their pain. a. Find a \(99 \%\) confidence interval for the difference in the proportions experiencing relief from pain for these two pain relievers. b. Based on the confidence interval in part a, is there sufficient evidence to indicate a difference in the proportions experiencing relief for the two pain relievers? Explain.

Explain what is meant by "margin of error" in point estimation.
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