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Compute a \(98 \%\) confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives.

Short Answer

Expert verified
The \(98 \%\) confidence interval for the proportion of defective items is \([0.0348, 0.1252]\).

Step by step solution

01

Identify Given Values

The question provides several pieces of information: the sample size \(n\) is 100, the number of defective items \(x\) is 8, and the confidence level is \(98 \%\). Calculate the sample proportion \(p = \frac{x}{n} = \frac{8}{100} = 0.08\).
02

Compute z-value

We need to calculate the z-value that corresponds to the middle of \(98 \%\) of the distribution. Since we have a two-tail test, the total area we are leaving out of the distribution is \(100 \% - 98 \% = 2 \%\), and that is evenly divided between the two tails. Therefore, we are leaving an area of \(1 \%\) in each tail. This corresponds to a z-value of approximately 2.33 from a standard normal distribution table.
03

Calculate the Margin of Error

The margin of error (E) can be calculated using the formula: \(E = z * \sqrt{\frac{p(1-p)}{n}}\). Substituting our values into the formula gives us: \(E = 2.33 * \sqrt{\frac{0.08(1-0.08)}{100}} = 0.0452\).
04

Calculate Confidence Interval

The confidence interval is found by taking the sample proportion \(\pm\) the margin of error. Therefore our \(98 \%\) confidence interval for the proportion of defective items is: \(0.08 \pm 0.0452\), or in interval notation \([0.0348, 0.1252]\).

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Key Concepts

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

Sampling Distribution
The concept of a sampling distribution is central to the field of statistics. It refers to the probability distribution of a given statistic based on a random sample. When we collect data through samples, we can think of each sample as providing us with an estimate of a population parameter, such as the mean or proportion. However, different samples can give different estimates. If we were to repeat the sampling process many times, we'd end up with a distribution of those estimates called the sampling distribution.

The characteristics of a sampling distribution, like its mean and standard deviation, are influenced by the size of the sample and the variation in the population. For a large enough sample size, thanks to the Central Limit Theorem, the sampling distribution of the sample mean tends to be normally distributed regardless of the shape of the population distribution. This is why we often rely on the normal distribution for creating confidence intervals, like in the exercise we are discussing.
Margin of Error
Margin of error represents how much we can expect a survey result to reflect the true population value. To put it simpler, it's an expression of the amount of random sampling error in the results. It gives us a range within which we can be certain that the population parameter lies, according to a specific confidence level.

Formulaically, it's calculated using the standard deviation of the sampling distribution or standard error, and the desired confidence level, which translates into the z-value. In real-world scenarios, such as political polling or quality control, understanding margin of error helps stakeholders make informed decisions. It's important to note that the margin of error decreases with increased sample size or less variation within the data, leading to more precise estimates.
Z-Value
The z-value, also known as a z-score, is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. In the context of confidence intervals, the z-value determines how far out from the mean we must go to capture the central portion of the normal distribution corresponding to the desired confidence level.

For example, for a 98% confidence level, we target the central 98% of the standard normal distribution and find the z-value that leaves just 1% in each tail, as demonstrated in the step-by-step solution where we used a z-value of approximately 2.33. This z-value is then used in the formula for margin of error, impacting how wide the confidence interval is. The further out we go (a higher z-value), the wider the interval, accommodating more uncertainty, but increasing our confidence that the interval includes the true population parameter.

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Most popular questions from this chapter

A random sample of 30 firms dealing in wireless products were selected to determine the proportion of such firms that have implemented new software to improve productivity. It turned out that eight of the 30 have implemented such software. Find a \(95 \%\) confidence interval on \(p\), the true proportion of such firms that have implemented new software.

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand \(A\) had an average: tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand \(B\) had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a \(95 \%\) confidence interval for the difference of the population means.

A machine is producing metal pieces that are cylindrical in shape. A sample of pieces is taken and the diameters are 1.01,0.97,1.03,1.04,0.99,0.98,0.99 \(1.01,\) and 1.03 centimeters. Find a \(99 \%\) confidence interval for the mean diameter of pieces from this machine, assuming an approximate normal distribution.

Regular consumption of presweetened cereals contributes to tooth decay, heart disease, and other degenerative diseases, according to studies conducted by Dr. W. H. Bowen of the National Institute of Health and Dr. J. Yudben, Professor of Nutrition and Dietetics at the University of London. In a random sample of 20 similar single servings of Alpha-Bits, the average sugar content was 11.3 grams with a standard deviation of 2.45 grams. Assuming that the sugar contents are normally distributed, construct a \(95 \%\) confidence interval for the mean sugar content for single servings of Alpha-Bits.

A new rocket-launching system is being considered for deployment of small, short-range rockets. The existing system has \(p=0.8\) as the probability of a successful launch. A sample of 40 experimental launches is made with the new system and 34 are successful. (a) Construct a \(95 \%\) confidence interval for \(p\). (b) Would you conclude that the new system is better?

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