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In the manufacture of a food product, the label states the box contains \(10 \mathrm{lbs}\). The boxes are filled by machine and it is of interest to determine whether or not the machine is set properly. Suppose the standard deviation is approximately .05, though this not precisely known. The company feels that the present setting is unsatisfactory if the machine fills the boxes so that the mean weight differs from \(10 \mathrm{lbs}\). by more than \(.01 \mathrm{lb}\). and it wants to be able to detect such a difference at least \(95 \%\) of the time. At a level of significance of \(\alpha=.05\), what sample size is needed?

Short Answer

Expert verified
To detect a mean weight difference of more than \(0.01 \mathrm{lb}\) with a \(95\%\) level of confidence and a level of significance of \(\alpha = 0.05\), a sample size of \(n = 97\) boxes is needed.

Step by step solution

01

Find the Z-score of the level of significance

Since the level of significance is \(\alpha = 0.05\), we need to find the Z-score corresponding to the tail area of \(0.025\) (two-tailed test). We use the Z-table or calculator to find this value, which is: \(Z_{\alpha/2}=1.96\).
02

Calculate the margin of error

The margin of error that the company wants to detect is \(.01 \mathrm{lb}\). We denote this value as \(E\), thus \(E = 0.01\).
03

Calculate the sample size

Now, we can use the formula given above to find the required sample size: \[n = \left(\frac{Z_{\alpha/2}\sigma}{E}\right)^2\] Plugging in the values, we get: \[n = \left(\frac{1.96 \times 0.05}{0.01}\right)^2\] \[n = \left(\frac{0.098}{0.01}\right)^2\] \[n \approx 96.04\] Since the sample size must be a whole number, we round up to the nearest whole number, as doing so will increase the level of confidence: \[n = 97\]
04

Conclusion

To detect a difference in the mean weight of more than \(.01 \mathrm{lb}\) with a \(95\%\) level of confidence and a level of significance of \(\alpha = 0.05\), the company needs a sample size of \(n = 97\) boxes.

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Key Concepts

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

Level of Significance
The "Level of Significance" is a core element in statistical hypothesis testing. It represents the threshold of risk we're willing to accept when making claims about the population, based on a sample. In statistical terms, it's denoted by \( \alpha \), and often set at \( 0.05 \), which corresponds to a 5% risk of mistakenly rejecting the true null hypothesis. In our scenario, where we're assessing whether our filling machine is correctly calibrated, setting \( \alpha = 0.05 \) means we're comfortable with a 5% chance of an error in detecting a machine inaccuracy when there actually isn't one. This careful balancing act of risk helps ensure robust decision-making in industrial settings.
Z-score
A "Z-score" is a way of quantifying the position of a data point relative to the mean of a group of values. This statistical measure indicates how many standard deviations an element is from the mean. Practically, it helps us understand how unusual or typical a sample observation is. In hypothesis testing, Z-scores play a critical role. For a two-tailed test like the one described, where the level of significance is \(0.05\), the corresponding Z-score is \(1.96\). This value divides the standard normal distribution so that 95% of data lies between \(-1.96\) and \(1.96\), thus defining the critical region where differences might be deemed statistically significant. This understanding helps us gauge the effectiveness and precision of the machine's settings.
Margin of Error
The "Margin of Error" is pivotal when evaluating the precision of a sample estimate. It expresses the extent of allowable error in the results, representing the range within which the true population parameter is expected to lie. In our example, the margin of error is \(0.01 \) lb. This means we are setting the boundary for how much the average weight of the machine-filled boxes can deviate from the target 10 lbs. When we compute the sample size, this margin ensures we are capable of detecting such a minimal deviation reliably. Thus, the margin of error aligns the statistical investigation with practical requirements, ensuring that if any significant deviation occurs, it will likely be noticed through the sample.
Standard Deviation
"Standard Deviation" is a crucial statistical metric that quantifies the degree of dispersion in a dataset. It indicates how much individual data points deviate, on average, from the mean. In quality control operations, like monitoring a machine's output, a small standard deviation (~0.05 lbs in this case) implies the weights of the boxes are mostly consistent. Understanding this helps in determining the necessary sample size to confidently identify any deviations from the expected norm. The formula for the required sample size embodies standard deviation, scaling it by the Z-score and dividing by the margin of error. This ensures that the sample sufficiently accounts for variability, allowing us to draw meaningful conclusions from the sample data related to the performance of the machinery.

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Most popular questions from this chapter

Suppose you are a buyer of large supplies of light bulbs. You want to test, at the \(5 \%\) significance level, the manufacturer's claim that his bulbs last more than 800 hours. You test 36 bulbs and find that the sample mean, \(\underline{X}\), is 816 hours and the sample standard deviation \(\mathrm{s}=70\) hours. Should you accept the claim?

Suppose a coin is assumed to be biased and it is believed that the probability, p, of obtaining a head on one toss is \(1 / 10\). Suppose the coin is tossed twice. Is there any value for \(\mathrm{k}\), the number of heads; for which we can reject the null hypothesis \(\mathrm{p}=1 / 10\) in favor of the alternate hypothesis \(\mathrm{p}>1 / 10\) at the \(1 \%\) level of significance? Suppose the null hypothesis is \(p=1 / 2\). Is there any value, \(k\), for which we can reject this null hypothesis in favor of the alternate hypothesis \(p<1 / 2\) at the \(1 \%\) level of significance when the coin is tossed twice?

In one income group, \(45 \%\) of a random sample of people express approval of a product. In another income group, \(55 \%\) of a random sample of people express approval. The standard errors for these percentages are \(.04\) and \(.03\) respectively. Test at the \(10 \%\) level of significance the hypothesis that the percentage of people in the second income group expressing approval of the product exceed that for the first income group.

The standard deviation of a particular dimension of a metal component is small enough so that it is satisfactory in subsequent assembly. A new supplier of metal plate is under consideration and will be preferred if the standard deviation of his product is not larger, because the cost of his product is lower than that of the present supplier. A sample of 100 items from each supplier is obtained. The results are as follows: New supplier: \(\mathrm{S}_{1}^{2}=.0058\) Old supplier: \(\mathrm{S}_{2}{ }^{2}=.0041\) Should the new supplier's metal plates be purchased? Test at the \(5 \%\) level of significance.

A certain printing press is known to turn out an average of 45 copies a minute. In an attempt to increase its output, an alteration is made to the machine, and then in 3 short test runs it turns out 46,47, and 48 copies a minute. Is this increase statistically significant, or is it likely to be simply the result of chance variation? Use a significance level of \(.05\).

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