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In Chapter 20, we saw that to construct a confidence interval for a population proportion it was enough to know the sample proportion and the sample size. Is the same true for constructing a confidence interval for a population mean? That is, is it enough to know the sample mean and sample size? Explain.

Short Answer

Expert verified
No, constructing a confidence interval for a population mean requires the sample standard deviation.

Step by step solution

01

Understanding Confidence Intervals for Proportions

In constructing a confidence interval for a population proportion, we use the sample proportion (\( \hat{p} \)) and the sample size (\( n \)). The formula for the confidence interval is given by \( \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( Z \) is the critical value from the standard normal distribution.
02

Exploring Confidence Intervals for Means

For a population mean, the confidence interval uses the sample mean (\( \bar{x} \)), the sample standard deviation (\( s \)), and the sample size (\( n \)). The formula is \( \bar{x} \pm t \times \frac{s}{\sqrt{n}} \), where \( t \) is the critical value from the t-distribution, which accounts for the sample size and variability.
03

Identifying Additional Information Needed

Unlike the proportion, constructing a confidence interval for a mean requires knowledge of the sample standard deviation, not just the sample mean and size. This is because the standard deviation measures the variability or spread of the data.
04

Conclusion on Information Requirements

Therefore, it is not enough to only know the sample mean and sample size for constructing a confidence interval for a population mean. The standard deviation is essential for determining the width of the confidence interval, reflecting the data variability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Mean
When talking about the population mean, we're focused on finding the average value for a whole group. In statistics, this group is often too large to measure entirely, so we use a sample, which is a smaller section of the population.
The sample mean (\( \bar{x} \)) acts as an estimate for the true population mean, giving us a glimpse into what the average might be for the entire population. However, unlike the sample proportion, knowing just the sample mean isn't sufficient to construct a reliable confidence interval for the population mean.
This is because the sample mean alone doesn't account for the data's spread or variability. Confidence intervals provide a range where we expect the true population mean to lie. This interval depends heavily on understanding how spread out the data points are, which we measure using the sample's standard deviation.
Population Proportion
When we determine the population proportion, we're measuring what fraction or percentage of the population has a certain characteristic. In situations where we’re dealing with large groups or populations, we take samples to make estimations.
To construct a confidence interval for a population proportion, you only need the sample proportion (\( \hat{p} \)) and sample size (\( n \)). The formula typically used is:
  • \[ \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Here, \( Z \) is the critical value from the Z-distribution, depicting how data is scattered around the average. This method doesn't require information on variability beyond the sample proportion and size, making it a bit simpler than the method for population means.
Sample Standard Deviation
Sample standard deviation (\( s \)) measures how much individual data points differ from the sample mean. It's a crucial element for constructing a confidence interval for a population mean because it reflects the variability in your data.
In simple terms, the sample standard deviation informs us about the spread of data. A small standard deviation means data points are close to the mean, while a larger one indicates more spread out data.
When forming a confidence interval to estimate the population mean, this variability is fundamental because it affects the confidence interval's width. Larger variability results in a wider confidence interval, reflecting more uncertainty in pinpointing the exact population mean.
T-distribution
The T-distribution is used in statistics to estimate population parameters when the sample size is small or the population standard deviation is unknown. The T-distribution resembles the Z-distribution (a normal distribution), but it has heavier tails, meaning it accounts for more variability or unpredictability in data estimates from small samples.
When constructing confidence intervals for a population mean, we employ the T-distribution to find the critical value (\( t \)), which adjusts according to the sample size's degrees of freedom (\( n-1 \)).
  • This use of degrees of freedom adjusts the shape of the T-distribution to address potential inaccuracies.
Overall, the T-distribution provides a more cautious estimate, offering more reliable confidence intervals with limited or variable data.
Z-distribution
The Z-distribution, or standard normal distribution, is a critical concept in constructing confidence intervals, especially for population proportions. The Z-distribution is symmetric and bell-shaped, centered around a mean of 0 and a standard deviation of 1.
It provides us with critical values (\( Z \)) which are used to calculate the margin of error for confidence intervals. When the sample size is large (typically \( n > 30 \)), or when dealing with population proportions, the Z-distribution is appropriate as the data tends to approximate normal distribution well.
In confidence intervals for population proportions, the Z-distribution simplifies assumptions, allowing statisticians to estimate the range within which the true proportion likely falls. Its use is limited when dealing with means from small samples, where instead, the T-distribution is a better fit.

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Most popular questions from this chapter

In a study comparing age of death for left- and right-handed baseball players, Coren and Halpern (1991, p. 93) provided the following information: "Mean age of death for strong right-handers was 64.64 years (SD = 15.5, n = 1472); mean age of death for strong left-handers [was] 63.97 years ( SD \(=15.4, n=236\) )" The term "strong handers" applies to baseball players who both threw and batted with the same hand. The data were actually taken from entries in The Baseball Encyclopedia (6th ed., New York: Macmillan, 1985), but, for the purposes of this exercise, pretend that the data were from a sample drawn from a larger population. a. Compute an approximate \(95 \%\) confidence interval for the mean age of death for the population of strong right-handers from which this sample was drawn. b. Repeat part (a) for the strong left-handers. c. Compare the results from parts (a) and (b) in two ways. First, explain why one confidence interval is substantially wider than the other. Second, explain whether you would conclude that there is a difference in the mean ages of death for left-and right-handers on the basis of these results. d. Compute an approximate \(95 \%\) confidence interval for the difference in mean ages of death for the strong right-and left-handers. Interpret the result.

Suppose you were given a \(95 \%\) confidence interval for the relative risk of disease under two different conditions. What could you conclude about the risk of disease under the two conditions if a. The confidence interval did not cover \(1.0 .\) b. The confidence interval did cover 1.0.

Suppose you were given a \(95 \%\) confidence interval for the difference in two population means. What could you conclude about the population means if a. The confidence interval did not cover zero. b. The confidence interval did cover zero.

Using the data presented by Hand and colleagues (1994) and discussed in previous chapters, we would like to estimate the average age difference between husbands and wives in Britain. Recall that the data consisted of a random sample of 200 couples. Following are two methods that were used to construct a confidence interval for the difference in ages. Your job is to figure out which method is correct: Method 1: Take the difference between the husband's age and the wife's age for each couple, and use the differences to construct an approximate \(95 \%\) confidence interval for a single mean. The result was an interval from 1.6 to 2.9 years. Method 2: Use the method presented in this chapter for constructing an approximate confidence interval for the difference in two means for two independent samples. The result was an interval from -0.4 to 4.3 years. Explain which method is correct, and why. Then interpret the confidence interval that resulted from the correct method.

The Baltimore Sun (Haney, 21 February 1995) reported on a study by Dr. Sara Harkness in which she compared the sleep patterns of 6 -month-old infants in the United States and the Netherlands. She found that the 36 U.S. infants slept an average of just under 13 hours out of every \(24,\) whereas the 66 Dutch infants slept an average of almost 15 hours. a. The article did not report a standard deviation, but suppose it was 0.5 hour for each group. Compute the standard error of the mean (SEM) for the U.S. babies. b. Continuing to assume that the standard deviation is 0.5 hour, compute an approximate 95\% confidence interval for the mean sleep time for 6 -month-old babies in the United States.c. Continuing to assume that the standard deviation for each group is 0.5 hour, compute an approximate \(95 \%\) confidence interval for the difference in average sleep time for 6 -month-old Dutch and U.S. infants.

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