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The variation in time for a baked good to go through a conveyor oven at a large scale bakery has standard deviation 0.017 minute at every time setting. To check the bake time of the oven periodically four batches of goods are carefully timed. The recent check gave a mean of 27.2 minutes with sample standard deviation 0.012 minute. Construct a \(99.8 \%\) confidence interval for the mean bake time of all batches baked in this oven. Assume bake times are normally distributed. Hint: Not all the numbers given in the problem are used.

Short Answer

Expert verified
The 99.8% confidence interval is (27.15983, 27.24017) minutes.

Step by step solution

01

Determine the Known Values

We have a sample mean \( \overline{x} = 27.2 \) minutes, a sample standard deviation \( s = 0.012 \) minutes, a standard deviation for the entire process \( \sigma = 0.017 \) minutes, a sample size \( n = 4 \), and a confidence level of \( 99.8\% \). We note that the population standard deviation is given but not typically used in a t-distribution confidence interval.
02

Choose the Appropriate Distribution

Given the small sample size of \( n = 4 \), we should use the t-distribution to construct the confidence interval rather than the z-distribution, even though \( \sigma \) is given, because the t-distribution accounts for the additional uncertainty in estimating the population standard deviation from a small sample.
03

Find the t-score for a 99.8% Confidence Interval

For \( n - 1 = 3 \) degrees of freedom (because \( n = 4 \)), and a 99.8% confidence level, we refer to the t-distribution table to find the t-score. The critical t-score for a two-tailed interval of 99.8% with 3 degrees of freedom is approximately \( t^{*} = 6.695 \).
04

Calculate the Standard Error (SE)

The standard error of the mean for this sample is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.012}{\sqrt{4}} = 0.006 \] minutes.
05

Construct the Confidence Interval

The confidence interval is calculated using the formula: \[ \overline{x} \pm t^{*} \times SE \] Substituting the values, the margin of error (ME) is: \[ ME = 6.695 \times 0.006 = 0.04017 \] Thus, the confidence interval becomes: \[ 27.2 \pm 0.04017 = (27.15983, 27.24017) \] minutes.

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Key Concepts

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

t-distribution
When working with confidence interval estimation and small samples, it's crucial to understand the role of the t-distribution. This distribution comes into play when you have a small sample size, typically less than 30. In the given problem set, our sample size is just 4, which is definitely small.
The t-distribution is wider and has heavier tails compared to the normal distribution. This means it accounts for the extra variability present in small samples. The smaller the sample size, the more the t-distribution deviates from the normal distribution, giving more conservative estimates with wider confidence intervals.
  • Why t-distribution? It's used when sample sizes are small and/or when population standard deviation is unknown.
  • Degrees of freedom (df): It's usually calculated as the sample size minus one, so for our case, it's 3 (because we have 4 samples).
  • Critical value: We use the t-distribution table to find the critical t-score for our specified confidence level (99.8% here).
In our exercise, this t-distribution compensates for the smaller sample size, making our confidence interval estimation more accurate.
standard error
The standard error (SE) is another fundamental concept in the context of confidence intervals. It acts as an indicator of the typical distance that the sample mean would be expected to deviate from the true population mean. In simpler terms, it helps us understand the accuracy of the sample mean.
For a set of data points, the standard error is calculated using the formula:\[SE = \frac{s}{\sqrt{n}}\]
Where:
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
In our example, the sample standard deviation is 0.012 minutes, and the number of observations is 4. Plugging these values into our formula gives us a standard error of 0.006 minutes.
The standard error provides a foundation for building the confidence interval: the smaller the SE, the more precise our estimate, leading to a narrower confidence interval.
sample mean
The sample mean is a simple but critical concept. It's the average of all observations in your sample, serving as an estimate of the population mean. In this exercise, our sample mean was calculated as 27.2 minutes.
When calculating a confidence interval, the sample mean is at the center of the interval. Essentially, you're trying to see how far the true population mean might be from this sample mean, within a certain level of confidence.
  • It's the best estimate of the population mean based on the sample data.
  • Used as the central value of the confidence interval: \( \overline{x} \pm ME \).
For the problem at hand, the calculated sample mean of 27.2 minutes suggests that on average, a batch takes around this much time in the oven. The confidence interval constructed around the sample mean lets us express how confident we are that this sample mean is close to the actual population mean.

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Most popular questions from this chapter

Confidence intervals constructed using the formula in this section often do not do as well as expected unless \(n\) is quite large, especially when the true population proportion is close to either 0 or \(1 .\) In such cases a better result is obtained by adding two successes and two failures to the actual data and then computing the confidence interval. This is the same as using the formula $$ \begin{aligned} &\tilde{p} \pm z_{\alpha} / 2 \sqrt{\frac{\hat{p}(1-\dot{p})}{\tilde{n}}}\\\ &\text { where }\\\ &\tilde{p}-\frac{x+2}{n+4} \operatorname{and} n-n+4 \end{aligned} $$ Suppose that in a random sample of 600 households, 12 had no telephone service of any kind. Use the adjusted confidence interval procedure just described to form a \(99.9 \%\) confidence interval for the proportion of all households that have no telephone service of any kind.

The body mass index (BMI) was measured in 1,200 randomly selected adults, with the results shown in the table. a. Give a point estimate for the proportion of all men whose BMI is over 25 . b. Assuming the sample is sufficiently large, construct a \(99 \%\) confidence interval for the proportion of all men whose BMI is over 25 c. Give a point estimate for the proportion of all adults, regardless of gender, whose BMI is over \(25 .\) d. Assuming the sample is sufficiently large, construct a \(99 \%\) confidence interval for the proportion of all adults, regardless of gender, whose BMI is over \(25 .\)

The amount of a particular biochemical substance related to bone breakdown was measured in 30 healthy women. The sample mean and standard deviation were 3.3 nanograms per milliliter \((\mathrm{ng} / \mathrm{mL})\) and \(1.4 \mathrm{ng} / \mathrm{mL}\). Construct an \(80 \%\) confidence interval for the mean level of this substance in all healthy women.

A corporation that owns apartment complexes wishes to estimate the average length of time residents remain in the same apartment before moving out. A sample of 150 rental contracts gave a mean length of occupancy of 3.7 years with standard deviation 1.2 years. Construct a \(95 \%\) confidence interval for the mean length of occupancy of apartments owned by this corporation.

A thread manufacturer tests a sample of eight lengths of a certain type of thread made of blended materials and obtains a mean tensile strength of 8.2 Ib with standard deviation 0.06 lb. Assuming tensile strengths are normally distributed, construct a \(90 \%\) confidence interval for the mean tensile strength of this thread.

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