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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.

Short Answer

Expert verified
The 80% confidence interval is (52.944, 54.056).

Step by step solution

01

Identify the Known Values

First, let's identify the given values from the problem. We know:- The mean \( \bar{x} = 53.5 \)- The standard deviation \( s = 1.6 \)- The sample size \( n = 15 \)- The confidence level is \(80\%\).
02

Determine the t-value

Since the sample size is small (less than 30) and the population standard deviation is unknown, we use the t-distribution. The degrees of freedom \( df = n - 1 = 14 \). For an \(80\%\) confidence interval, the t-value can be found using a t-table or calculator, which is approximately \( t_{\alpha/2} = 1.345 \) for \( df = 14 \).
03

Calculate the Standard Error

The standard error (SE) of the mean is calculated using:\[ SE = \frac{s}{\sqrt{n}} = \frac{1.6}{\sqrt{15}} \approx 0.413 \]
04

Calculate the Margin of Error

The margin of error (ME) is calculated using the t-value and standard error:\[ ME = t_{\alpha/2} \times SE = 1.345 \times 0.413 \approx 0.556 \]
05

Construct the Confidence Interval

Finally, construct the confidence interval by adding and subtracting the margin of error from the mean:\[ CI = \bar{x} \pm ME = 53.5 \pm 0.556 \]This gives the interval:\[ CI = (53.5 - 0.556, 53.5 + 0.556) = (52.944, 54.056) \]

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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 creating a confidence interval and the sample size is small, generally less than 30, the t-distribution is often used. Unlike a normal distribution, the t-distribution is wider and accounts for the uncertainty that comes with estimating the population standard deviation from a small sample. It is specifically used when:
  • The sample size is small.
  • The population standard deviation is unknown.
The t-distribution is similar to the normal distribution but with thicker tails. These thicker tails provide more variability, so when estimating with smaller samples, you get more conservative estimates. It has a parameter called degrees of freedom, which is typically based on the sample size minus one (\( df = n - 1 \)).
This degree of freedom affects the exact shape of the t-distribution curve. As the sample size increases, the t-distribution approaches the normal distribution.
Decoding Margin of Error
The margin of error is an important aspect of constructing a confidence interval. It tells you how far your sample mean is likely to be from the actual population mean. It's calculated by multiplying the t-value with the standard error.
  • The t-value, as mentioned earlier, comes from the t-distribution and depends on your confidence level.
  • The standard error represents the standard deviation of the sample mean.
In simpler terms, the margin of error gives you a range, set at approximately \(1\) standard error on either side of the sample mean, allowing us to say, "we are this confident that the actual mean lies within this range."
This provides insight into the precision of your estimate, giving a balance between accuracy and confidence.
Introducing Standard Error
Standard error (SE) measures how much the sample mean is expected to vary from the true population mean. It is calculated by dividing the sample standard deviation by the square root of the sample size: \[ SE = \frac{s}{\sqrt{n}} \]
  • The standard deviation \(s\) tells us about the spread of observations in our sample.
  • The sample size \(n\) plays a critical role because, generally, the larger the sample, the closer the sample mean is to the population mean.
This is why larger samples tend to provide more accurate estimates. The standard error reflects this accuracy by becoming smaller as the sample size increases, indicating less variability in the sample mean. Understanding this concept is key in interpreting how confidently we can make inferences about the population from the sample.

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Most popular questions from this chapter

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.

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Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given. a. \(p=0.81,95 \%\) confidence, \(E=0.02\) b. \(\quad p=0.81,99 \%\) confidence, \(E=0.02\) c. \(\quad p=0.81,95 \%\) confidence, \(E=0.01\)

A sample of 250 workers aged 16 and older produced an average length of time with the current employer ("job tenure") of 4.4 years with standard deviation 3.8 years. Construct a \(99.9 \%\) confidence interval for the mean job tenure of all workers aged 16 or older.

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