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Chapter 8: Problem 13

The standard deviation for a population is \(\sigma=7.14\). A random sample selected from this population gave a mean equal to \(48.52\). a. Make a \(95 \%\) confidence interval for \(\mu\) assuming \(n=196\). b. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=100\). c. Determine a \(95 \%\) confidence interval for \(\mu\) assuming \(n=49\). d. Does the width of the confidence intervals constructed in parts a through c increase as the sample size decreases? Explain.

Short Answer

Expert verified
a. The 95% confidence interval for a sample size of 196 is [47.96, 49.08].\nb. The 95% confidence interval for a sample size of 100 is [47.76, 49.28].\nc. The 95% confidence interval for a sample size of 49 is [47.44, 49.60].\nd. Yes, the width of the confidence intervals increases as the sample size decreases, suggesting that the precision of the estimate decreases with smaller sample sizes.

Step by step solution

01

Find confidence intervals for different sample sizes

Plug the given mean (\(48.52\)), standard deviation (\(7.14\)), and sample sizes (\(n=196\), \(n=100\), and \(n=49\)) into the formula. This gives the following results:\na. For \(n=196\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{196}, 48.52 + 1.96 * 7.14/\sqrt{196}\)].\nb. For \(n=100\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{100}, 48.52 + 1.96 * 7.14/\sqrt{100}\)].\nc. For \(n=49\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{49}, 48.52 + 1.96 * 7.14/\sqrt{49}\)].
02

Compute the confidence intervals

Perform the mathematical calculations to find the confidence intervals for the different sample sizes. The results are:\na. For \(n=196\), the confidence interval is [\(47.96, 49.08\)].\nb. For \(n=100\), the confidence interval is [\(47.76, 49.28\)].\nc. For \(n=49\), the confidence interval is [\(47.44, 49.60\)].
03

Analyze the widths of the confidence intervals

Observe the widths of the confidence intervals for the different sample sizes. The width of a confidence interval is the difference between its upper and lower limits. We see that as the sample size decreases, the width of the confidence interval increases. This suggests that the precision of the estimate decreases as the sample size decreases.

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Key Concepts

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

Standard Deviation
Standard deviation is a vital statistical concept that measures how spread out numbers in a data set are. Imagine you took a test alongside your classmates. If most of you scored pretty close to the class average, the standard deviation would be small. But if scores were all over the place, from very low to very high, the standard deviation would be larger. Here's why it matters:
  • It helps understand the variability of your data.
  • A lower standard deviation means data points are close to the mean, while a higher one indicates more spread.
For this exercise, the population standard deviation is given as 7.14. This means that individual measurements fall within this range away from the average.
Understanding this helps us get a sense of how "typical" or "atypical" a data point might be.
Sample Size
Sample size plays a crucial role in statistics. It refers to the number of observations or data points you have collected. With the confidence interval problem at hand, the sample sizes are varied as 196, 100, and 49. Why does it matter?
  • A larger sample size reduces the margin of error in calculations.
  • Bigger samples tend to provide more reliable and accurate estimates of the population mean.
  • Smaller samples might lead to wider confidence intervals, meaning less precision.
In this exercise, notice how the confidence intervals change with different sample sizes. A larger sample (n = 196) gives a tighter interval, [47.96, 49.08], compared to a smaller sample (n = 49) with a wider interval, [47.44, 49.60].
This contrast shows how much your results can vary, hinging on how much data you gather.
Population Mean
The population mean is a central concept in statistics, representing the average of a set of data points for an entire group. In our case, the sample mean is 48.52. What's the difference between a sample mean and a population mean?
  • Sample mean is the average of a subset drawn from the population.
  • Population mean is the hypothetical average calculated if you had all data points from the population.
The sample mean serves as an estimate of the population mean. That's because, practically, it's often impossible to measure every single individual in the population.
Calculations with the sample mean help us draft a confidence interval, giving us a range where we believe the true population mean lies, with a certain level of confidence (like 95%). It's like saying, "We're 95% sure the actual average is somewhere around here."
Statistical Precision
Statistical precision refers to how much confidence we can have in our results. It's about how "tight" or "narrow" our estimates are. When we see terms like `confidence interval`, this precision is being evaluated.
For this exercise, what we've learned is:
  • The width of a confidence interval indicates precision. The narrower, the more precise.
  • Larger sample sizes generally lead to more precise estimates, evidenced by narrower confidence intervals.
  • Precision decreases as sample size decreases, making results less reliable.
The shift in confidence intervals in the provided solutions, depending on the sample size, reflects changes in statistical precision.
Practically, when precision is high, decisions based on this data are more likely to be accurate. This understanding helps in planning experiments or surveys effectively, ensuring that conclusions are strong and dependable.

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Most popular questions from this chapter

A gas station attendant would like to estimate \(p\), the proportion of all households that own more than two vehicles. To obtain an estimate, the attendant decides to ask the next 200 gasoline customers how many vehicles their households own. To obtain an estimate of \(p\), the attendant counts the number of customers who say there are more than two vehicles in their households and then divides this number by \(200 .\) How would you critique this estimation procedure? Is there anything wrong with this procedure that would result in sampling and/or nonsampling errors? If so, can you suggest a procedure that would reduce this error?

Yunan Corporation produces bolts that are supplied to other companies. These bolts are supposed to be 4 inches long. The machine that makes these bolts does not produce each bolt exactly 4 inches long but the length of each bolt varies slightly. It is known that when the machine is working properly, the mean length of the bolts made on this machine is 4 inches. The standard deviation of the lengths of all bolts produced on this machine is always equal to \(.04\) inch. The quality control department takes a random sample of 20 such bolts every week, calculates the mean length of these bolts, and makes a \(98 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(4.02\) inches or the lower limit of this confidence interval is less than \(3.98\) inches, the machine is stopped and adjusted. A recent such sample of 20 bolts produced a mean length of \(3.99\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the population distribution is approximately normal.

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=50 \quad\) and \(\quad \hat{p}=.25\) b. \(n=160\) and \(\hat{p}=.03\) c. \(n=400\) and \(\hat{p}=.65\) d. \(n=75 \quad\) and \(\quad \hat{p}=.06\)

The following data give the one-way commuting times (in minutes) from home to work for a random sample of 30 workers. \(\begin{array}{llllllllll}23 & 17 & 34 & 26 & 18 & 33 & 46 & 42 & 12 & 37 \\ 44 & 15 & 22 & 19 & 28 & 32 & 18 & 39 & 40 & 48 \\ 25 & 36 & 23 & 39 & 42 & 46 & 29 & 17 & 24 & 31\end{array}\) a. Calculate the value of the point estimate of the mean one-way commuting time from home to work for all workers. b. Construct a \(99 \%\) confidence interval for the mean one-way commuting time from home to work for all workers.

A random sample of 34 participants in a Zumba dance class had their heart rates measured before and after a moderate 10 -minute workout. The following data correspond to the increase in each individual's heart rate (in beats per minute): \(\begin{array}{llllllllllll}59 & 70 & 57 & 42 & 57 & 59 & 41 & 54 & 44 & 36 & 59 & 61 \\ 52 & 42 & 41 & 32 & 60 & 54 & 52 & 53 & 51 & 47 & 62 & 62 \\ 44 & 69 & 50 & 37 & 50 & 54 & 48 & 52 & 61 & 45 & & \end{array}\) a. What is the point estimate of the corresponding population mean? b. Make a \(98 \%\) confidence interval for the average increase in a person's heart rate after a moderate 10 -minute Zumba workout.
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