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The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?

Short Answer

Expert verified
The researcher should assume that the customer satisfaction index scores are normally distributed. The margin of error for a 95% confidence interval is 3.09, while for a 99% confidence interval, it is 4.079.

Step by step solution

01

Assumption for Margin of Error

The researcher should be willing to assume that the customer satisfaction index scores for the population are normally distributed to use the given population standard deviation \( (\sigma=5) \) and use a t-distribution or z-distribution for calculating the margin of error.
02

Margin of Error for 95% Confidence Interval

To calculate the margin of error for 95% confidence interval, we will use the formula: \[Margin\_of\_Error = z_{\alpha/2} * \frac{\sigma}{\sqrt{n}} \] Where \(z_{\alpha/2}\) is the z-score corresponding to the desired level of confidence, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Since we are given a 95% confidence level, we must find the \(z_{\alpha/2}\) score for 95%. This corresponds to the z-score of 1.96: \(z_{\alpha/2} = 1.96\) Now, we have everything to calculate the margin of error: \(Margin\_of\_Error = 1.96 * \frac{5}{\sqrt{10}}\) \(Margin\_of\_Error = 3.09\) The margin of error for a 95% confidence interval is 3.09.
03

Margin of Error for 99% Confidence Interval

To calculate the margin of error for 99% confidence interval, we will use the same formula: \[Margin\_of\_Error = z_{\alpha/2} * \frac{\sigma}{\sqrt{n}} \] This time, we are given a 99% confidence level, so we must find the \(z_{\alpha/2}\) score for 99%. This corresponds to the z-score of 2.576: \(z_{\alpha/2} = 2.576\) Now, we have everything to calculate the margin of error: \(Margin\_of\_Error = 2.576 * \frac{5}{\sqrt{10}}\) \(Margin\_of\_Error = 4.079\) The margin of error for a 99% confidence interval is 4.079.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that is used to estimate the true value of a population parameter. When we calculate a confidence interval, we are trying to say we are "confident" that the true value lies within this interval. For example, in the context of the exercise, the confidence interval gives an estimated range for the true customer satisfaction index for all fast-food or pizza group restaurants.
To construct a confidence interval, we need to know two key pieces of data: the sample mean and the margin of error. A wider interval may mean we are more confident it contains the true population parameter, while a narrower one may speak to higher precision of our estimate.
Normal Distribution
Normal distribution, also known as Gaussian distribution, is a very important concept in statistics. It's a continuous probability distribution that is symmetrical around the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean.
If we plot it, you will see it forming a bell-shaped curve. In the exercise, the assumption is made that the customer satisfaction index follows a normal distribution. This is important, as it allows us to use z-scores to calculate the confidence interval. The normality assumption is crucial for statistical validity in calculating the margin of error.
Sample Mean
The sample mean is the average value from a sample, calculated by summing all the observations and dividing by the number of observations. It is used as an estimate of the population mean, which is the true average of the entire population.
In our exercise, the sample mean of the customer satisfaction index is given as 71. This means the average satisfaction score from the sampled restaurants is 71.
Using the sample mean provides us an insight into the potential mean of the entire population, especially when combined with its measure of uncertainty, like the margin of error.
Population Standard Deviation
The population standard deviation is a crucial factor when estimating the variance of a dataset. It measures the dispersion or spread of the data relative to its mean. A larger standard deviation suggests more spread out data. By contrast, a smaller standard deviation points to data being clustered closely around the mean.
In this exercise, the standard deviation of the customer satisfaction index is given as a constant value across the population, \(\sigma = 5\). This consistent value enables us to create accurate confidence intervals, as it helps determine the margin of error. Moreover, having a known population standard deviation is often required to apply the normal distribution properties in inferential statistics.

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