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Chapter 7: Problem 16

A sample of 26 women's size 6 dresses had mean waist measurement 25.25 inches with sample standard deviation 0.375 inch. Construct a \(95 \%\) confidence interval for the mean waist measurement of all size 6 women's dresses. Assume waist measurements are normally distributed.

Short Answer

Expert verified
The 95% confidence interval for the mean waist measurement is approximately [25.10, 25.40] inches.

Step by step solution

01

Identify Known Values

We are given the sample mean as \( \bar{x} = 25.25 \) inches, the sample standard deviation as \( s = 0.375 \) inch, and the sample size \( n = 26 \). We need to construct a 95% confidence interval for the mean waist measurement.
02

Determine the Critical Value

Since the sample size is less than 30, we use the t-distribution. The degrees of freedom are \( n-1 = 25 \). For a 95% confidence interval, we look up the critical value \( t^* \) for 25 degrees of freedom, which is approximately 2.060.
03

Calculate the Standard Error of the Mean

The standard error (SE) is calculated using the formula: \( SE = \frac{s}{\sqrt{n}} \). Plugging in the values, we get \( SE = \frac{0.375}{\sqrt{26}} \approx 0.0735 \).
04

Calculate Margin of Error

The margin of error (ME) is given by \( ME = t^* \times SE \). Substituting the known values, \( ME = 2.060 \times 0.0735 \approx 0.1515 \).
05

Construct the Confidence Interval

The 95% confidence interval is calculated as \( \bar{x} \pm ME \). Hence, the interval is \( 25.25 \pm 0.1515 \), which is \( [25.0985, 25.4015] \).

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Key Concepts

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

Understanding the t-distribution
When we want to estimate a population mean from a small sample size, the t-distribution is often the best fit. This distribution is like the normal distribution but has heavier tails, which means it is more prone to producing values that fall far from the mean.
  • This heavy-tailed nature makes it ideal for smaller samples, usually those that contain fewer than 30 observations.
  • We use degrees of freedom (df), calculated as sample size minus one (\( n - 1 \)), to find the appropriate t-value used in calculations.
In our exercise, with \( n = 26 \), the degrees of freedom are 25. We find from the t-distribution table a t-value of approximately 2.060 for a 95% confidence interval. This value helps us determine how wide our interval will be, ensuring it accurately reflects the uncertainty in our estimate.
The Role of Standard Error
Standard error (SE) is a critical part of confidence interval calculations, as it gives us an estimate of how far the sample mean is likely to be from the true population mean.
  • Calculated using the formula \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size.
  • The smaller the standard error, the closer the sample mean likely is to the population mean.
In our problem with a sample standard deviation \( s = 0.375 \) and sample size \( n = 26 \), the SE computed is \( 0.0735 \). It signifies the measure of variability or dispersion of the sample mean from the actual population mean.
Calculating the Margin of Error
The margin of error (ME) works to quantify the extent to which the sample mean might deviate from the true population mean. This is calculated by multiplying the critical t-value by the standard error, \( ME = t^* \times SE \).
  • The critical t-value is derived from the t-distribution, as discussed previously.
  • The standard error gives us a sense of variability in our sample mean.
For our exercise, with the critical value \( t^* \) of approximately 2.060 and the earlier computed standard error of \( 0.0735 \), the margin of error calculated is \( ME = 0.1515 \). This means that our sample mean of 25.25 inches could differ from the true population mean by about 0.1515 inches in both directions, providing a range for our confidence interval.

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Most popular questions from this chapter

For all settings a packing machine delivers a precise amount of liquid; the amount dispensed always has standard deviation 0.07 ounce. To calibrate the machine its setting is fixed and it is operated 50 times. The mean amount delivered is 6.02 ounces with sample standard deviation 0.04 ounce. Construct a \(99.5 \%\) confidence interval for the mean amount delivered at this setting. Hint: Not all the information provided is needed.

A manufacturer of chokes for shotguns tests a choke by shooting 15 patterns at targets 40 yards away with a specified load of shot. The mean number of shot in a 30 -inch circle is 53.5 with standard deviation 1.6. Construct an \(80 \%\) confidence interval for the mean number of shot in a 30 -inch circle at 40 yards for this choke with the specified load. Assume a normal distribution of the number of shot in a 30 - inch circle at 40 yards for this choke.

A power wrench used on an assembly line applies a precise, preset amount of torque; the torque applied has standard deviation 0.73 foot-pound at every torque setting. To check that the wrench is operating within specifications it is used to tighten 100 fasteners. The mean torque applied is 36.95 foot-pounds with sample standard deviation 0.62 foot-pound. Construct a \(99.9 \%\) confidence interval for the mean amount of torque applied by the wrench at this setting. Hint: Not all the information provided is needed.

The designer of a garbage truck that lifts roll-out containers must estimate the mean weight the truck will lift at each collection point. A random sample of 325 containers of garbage on current collection routes yielded \(x-=75.3 \mathrm{lb}, \mathrm{s}=12.8 \mathrm{lb}\). Construct a \(99.8 \%\) confidence interval for the mean weight the trucks must lift each time.

Wildlife researchers tranquilized and weighed three adult male polar bears. The data (in pounds) are: 926,742,1,109 . Assume the weights of all bears are normally distributed. a. Construct an \(80 \%\) confidence interval for the mean weight of all adult male polar bears using these data. b. Convert the three weights in pounds to weights in kilograms using the conversion \(1 \mathrm{lb}=0.453 \mathrm{~kg}\) (so the first datum changes to \((\mathrm{g} 26)(0.453)-410)\). Use the converted data to construct an \(80 \%\) confidence interval for the mean weight of all adult male polar bears expressed in kilograms. c. Convert your answer in part (a) into kilograms directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.
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