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Chapter 8: Problem 105

A quality-control engineer wants to estimate the fraction of defectives in a large lot of printer ink cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a \(95 \%\) confidence interval?

Short Answer

Expert verified
Answer: The approximate sample size needed is 1824 printer ink cartridges.

Step by step solution

01

Identify the known values

From the problem, we know the following: - The estimated fraction of defectives (\(\hat{p}\)) is around 0.05. - The desired margin of error (\(E\)) is 0.01. - The confidence level is 95%, which corresponds to a z-score (\(z\)) of approximately 1.96.
02

Use the sample size formula

The sample size formula for proportions is given by: \(n = \frac{z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2}\) We'll plug in the known values and solve for the sample size (\(n\)).
03

Substitute the known values

Substituting the values we know into the formula: \(n = \frac{(1.96)^2 \cdot 0.05 \cdot (1 - 0.05)}{(0.01)^2}\)
04

Calculate the sample size

Now, we just need to perform the calculations: \(n = \frac{(3.8416) \cdot 0.05 \cdot (0.95)}{0.0001}\) \(n \approx 1823.82\) Since we can't have a fraction of a sample size, we need to round up to the nearest whole number.
05

Determine the required sample size

Thus, the quality-control engineer needs to take a sample size of approximately 1824 printer ink cartridges to estimate the true fraction to within .01, using a 95% confidence interval.

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Key Concepts

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

Confidence Interval
When we talk about confidence intervals, we're discussing a range that we believe contains the true value of a proportion. This range is calculated from the sample data. In essence, it gives us an estimation boundary. A 95% confidence interval means we're 95% confident that the true parameter lies within this range.
To compute a confidence interval for a sample proportion, we use the formula:
  • The estimated proportion (\(\hat{p}\)): The proportion we observed in our sample.
  • The z-score (\(z\)): A value from the z-distribution that corresponds to our confidence level.
  • The standard error: How much the sample proportion is expected to vary from the true proportion.
Combining these, the confidence interval for a proportion is expressed as \(\hat{p} \pm z \times \text{standard error}\).
This interval helps us understand the reliability of our estimate.
Margin of Error
The margin of error signifies how much we expect the true proportion to differ from our sample estimate. It's a measure of precision. In simple terms, it's an indication of the potential error we might face in our estimation.
The margin of error (\(E\)) is calculated using the formula:
  • \(E = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
This formula uses:
  • The z-score for the desired confidence level.
  • The sample proportion, which reflects the observed data.
  • The sample size.
When we have a smaller margin of error, our estimate is more precise. In the context of the original problem, the goal was to achieve a margin of error of 0.01. This implies the estimated proportion should not differ from the true proportion by more than 1%.
Proportion Estimation
Proportion estimation is all about figuring out the percentage or fraction of a particular trait in a population. In this case, the engineer wants to estimate the proportion of defective cartridges.
Ensuring an accurate estimate requires a good sample size. The formula used in the solution, \(n = \frac{z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2}\), helps determine this size. Here's a breakdown:
  • \(z^2\): Relates to how confident we want to be.
  • \(\hat{p}(1-\hat{p})\): The variability of the sample.
  • \(E^2\): The desired margin of error.
Plugging values into the formula gives us the number of samples needed to achieve our confidence level and margin of error. For instance, with an estimated defect proportion of 0.05 and a margin of error of 0.01, the engineer requires about 1824 samples to ensure a confident estimate.

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Most popular questions from this chapter

An experiment was conducted to estimate the effect of smoking on the blood pressure of a group of 35 cigarette smokers, by taking the difference in the blood pressure readings at the beginning of the experiment and again 5 years later. The sample mean increase, measured in millimeters of mercury, was \(\bar{x}=9.7,\) and the sample standard deviation was \(s=5.8\). Estimate the mean increase in blood pressure that one would expect for cigarette smokers over the time span indicated by the experiment. Find the margin of error. Describe the population associated with the mean that you have estimated.

In an Advertising Age white paper concerning the changing role of women as "breadwinners" in the American family, it was reported that according to their survey with JWT, working men reported doing 54 minutes of household chores a day, while working women reported tackling 72 minutes daily. But when examined more closely, Millennial men reported doing just as many household chores as the average working women, 72 minutes, compared to an average of 54 minutes among both Boomer men and Xer men. \({ }^{6}\) The information that follows is adapted from these data and is based on random samples of 1136 men and 795 women. a. Construct a \(95 \%\) confidence interval for the average time all men spend doing household chores. b. Construct a \(95 \%\) confidence interval for the average time women spend doing household chores.

Based on repeated measurements of the iodine concentration in a solution, a chemist reports the concentration as \(4.614,\) with an "error margin of .006." a. How would you interpret the chemist's "error margin"? b. If the reported concentration is based on a random sample of \(n=30\) measurements, with a sample standard deviation \(s=.017,\) would you agree that the chemist's "error margin" is .006?

Exercise 8.108 presented statistics from a study of fast starts by ice hockey skaters. The mean and standard deviation of the 69 individual average acceleration measurements over the 6 -meter distance were 2.962 and .529 meters per second, respectively. a. Find a \(95 \%\) confidence interval for this population mean. Interpret the interval. b. Suppose you were dissatisfied with the width of this confidence interval and wanted to cut the interval in half by increasing the sample size. How many skaters (total) would have to be included in the study?

Samples of 400 printed circuit boards were selected from each of two production lines \(A\) and \(B\). Line A produced 40 defectives, and line B produced 80 defectives. Estimate the difference in the actual fractions of defectives for the two lines with a confidence coefficient of \(.90 .\)
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