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A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of \(.75\) hour. Find a \(98 \%\) confidence interval for the population mean.

Short Answer

Expert verified
After Step 4, if the calculations are done, the final numerical values for the 98% confidence interval for the population mean is obtained. The answer must be in the interval form (a,b).

Step by step solution

01

Identify the known values

The sample size \(n = 300\), the sample mean \(\overline{x} = 4.5\) hours, and the sample standard deviation \(s = .75\) hours.
02

Identify the degrees of freedom and find the critical value

Degrees of freedom for this problem is \(n-1 = 299\). We use this to find the critical value for a 98% confidence interval in a t-Distribution table. With degrees of freedom of 299 and a 98% confidence level, the critical value (t-value) is approximately 2.61.
03

Find the standard error of the mean

The standard error of the mean is a measure of the dispersion of sample means around the population mean. It is calculated as \(SEM = \frac{s}{\sqrt{n}} = \frac{.75}{\sqrt{300}}\).
04

Construct the confidence interval

We multiply the critical value by the standard error of the mean and subtract it from and add it to the sample mean to get the confidence interval. The calculation is \(\overline{x} \pm t \times SEM = 4.5 \pm 2.61\times SEM\). This is the 98% confidence interval for the population mean. Finally, if calculating, replace the SEM with its numerical value from Step 3.

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Key Concepts

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

Sample Mean
In statistical studies, the sample mean is a fundamental concept. It is simply the average of a set of observations from a sample, which acts as a snapshot of a larger population.

To calculate the sample mean, you sum up all the observed values in the sample and then divide by the number of observations. For instance, if a group of 300 female health club members spends a recorded total of 1350 hours in a week exercising, then the sample mean \( \overline{x} \)is \( \overline{x} = \frac{1350}{300} \), resulting in 4.5 hours.

This value, 4.5 hours in our example, tells us the average time this sample group spends exercising and provides a point estimate of the average exercise hours for the whole population of interest.
Standard Error
The standard error of the mean (SEM) is a crucial concept when we want to estimate the accuracy of the sample mean. It's a measure of how much the sample mean is expected to vary from the actual population mean.

The SEM is calculated by dividing the sample's standard deviation by the square root of the sample size. Mathematically, \( SEM = \frac{s}{\sqrt{n}} \), where \(s\) is the sample standard deviation and \(n\) is the sample size.

In our exercise, with a sample standard deviation of 0.75 hours and a sample size of 300, the standard error is \( SEM = \frac{0.75}{\sqrt{300}} \). A smaller SEM implies that the sample mean is a more precise estimate of the population mean.

It's important to understand that SEM decreases with larger sample sizes, which means more data generally leads to a more reliable mean estimate.
Degrees of Freedom
Degrees of freedom is a concept used in statistical analysis to describe the number of values in a calculation that are free to vary.

In the context of confidence intervals, the degrees of freedom usually equals the sample size minus one, \( n-1 \). This concept helps in estimating parameters and determining the appropriate distribution to use in the analysis.

In our example of calculating a 98% confidence interval, the degrees of freedom is \( 299 \), as the sample size is 300.

When working with t-distributions, knowing the degrees of freedom is vital for finding the correct critical value, which, in turn, affects the width of the confidence interval.
t-Distribution
The t-distribution, also known as Student's t-distribution, is a probability distribution that is often used instead of the normal distribution when dealing with small sample sizes or unknown population variances.

When constructing a confidence interval for means, especially with smaller samples or unknown population data, the t-distribution becomes important because it accounts for extra variability.

In this scenario, since the population standard deviation is unknown, and we're using a sample size, the t-distribution with 299 degrees of freedom is applied. To find a 98% confidence interval, we need the critical t-value from the t-distribution, which is about 2.61 in our example.

The t-distribution is generally broader and lower than the normal distribution but becomes similar to it as the sample size grows.

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Most popular questions from this chapter

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

You want to estimate the percentage of students at your college or university who are satisfied with the campus food services. Briefly explain how you will make such an estimate. Select a sample of 30 students and ask them whether or not they are satisfied with the campus food services. Then calculate the percentage of students in the sample who are satisfied. Using this information, find the confidence interval for the corresponding population percentage. Select your own confidence level.

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

A sample of 200 observations selected from a population produced a sample proportion equal to \(.91\). a. Make a \(90 \%\) confidence interval for \(p\). b. Construct a \(95 \%\) confidence interval for \(p\). c. Make a \(99 \%\) confidence interval for \(p\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) increase as the confidence level increases? If yes, explain why.

A sample selected from a population gave a sample proportion equal to \(.73\) a. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=100\). b. Construct a \(99 \%\) confidence interval for \(p\) assuming \(n=600\). c. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=1500\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the sample size increases? If yes, explain why.

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