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The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season was \(70 .\) A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the population mean. b. Does the 99 percent confidence interval include the value suggested by the NCAA? Interpret this result. c. Suppose you decided to switch from a 99 to a 95 percent confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?

Short Answer

Expert verified
a. (65.61, 71.59) b. Yes, it includes 70 hours. c. The interval will decrease.

Step by step solution

01

Identify the information

First, identify the given information from the problem:- Mean of sample (\( \bar{x} \)) = 68.6 hours- Standard deviation (\( s \)) = 8.2 hours- Sample size (\( n \)) = 50- Confidence level = 99%- Mean reported by NCAA = 70 hours.
02

Find critical value for confidence interval

For a 99% confidence interval, we need the critical value, which can be found using a standard normal (Z) distribution. This is because the sample size is large (\( n \geq 30 \)). The Z-critical value for a 99% confidence interval is approximately 2.576.
03

Use the formula for confidence interval

The formula for the confidence interval is: \[ \bar{x} \pm Z \left( \frac{s}{\sqrt{n}} \right) \]Where \( \bar{x} \) is the sample mean, \( Z \) is the critical value, \( s \) is the standard deviation, and \( n \) is the sample size.
04

Calculate the margin of error

Plug the values into the formula to calculate the margin of error:\[ 2.576 \times \left( \frac{8.2}{\sqrt{50}} \right) \approx 2.9913 \]
05

Construct the confidence interval

Add and subtract the margin of error from the sample mean:\[ 68.6 - 2.9913 = 65.6087 \]\[ 68.6 + 2.9913 = 71.5913 \]So, the 99% confidence interval for the population mean is (65.61, 71.59).
06

Interpret the confidence interval

Since 70 hours (NCAA's reported mean) falls within the 99% confidence interval (65.61, 71.59), we do not have enough evidence to say that the mean hours reported by the NCAA is incorrect.
07

Effect of changing confidence level

Switching from a 99% to a 95% confidence interval will decrease the interval width because the critical value will be smaller (approximately 1.96), thus the margin of error will be less.

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Key Concepts

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

Population Mean
When we talk about the population mean, we refer to the average value of a particular characteristic across an entire group or population. In statistics, we often wish to estimate this value when it is not feasible to measure the whole population directly. Instead, we rely on samples, small groups selected from the population, to make inferences about the population mean. For instance, in the given exercise, the NCAA claims that college football assistant coaches spend an average (population mean) of 70 hours per week on their duties. Understanding whether this reported mean is representative of all assistant coaches is crucial for accurate analysis and decision-making.
Critical Value
The concept of a critical value is central to constructing confidence intervals. It acts as a "cut-off" point that determines how extreme data points must be for them to be considered statistically significant. In the context of the exercise, the critical value is derived from the Z-distribution because our sample size is large; this is known as the Z-critical value. For a 99% confidence level, the critical value is approximately 2.576. This value is chosen to ensure that the confidence interval captures the population mean 99% of the time. Therefore, understanding the critical value helps assess the reliability of our interval estimation.
Margin of Error
The margin of error is an essential component of confidence intervals. It quantifies the extent of possible error in our sample estimate when inferring the population parameter. Essentially, it broadens the range around our sample mean to account for sampling variability and ensures a degree of confidence in our estimation. In calculating the margin of error for the given problem, we multiply the critical value by the standard deviation of the sample divided by the square root of the sample size. Using the given figures, the margin of error computes to around 2.9913 hours. This means the true population mean can realistically lie within a range of approximately 2.99 hours above or below our sample mean.
Sample Mean
The sample mean, often represented by \( \bar{x} \), is the average value calculated from a sample that provides an estimate of the population mean. It's a pivotal starting point for any statistical analysis aimed at inferencing population characteristics.For the situation posed by the exercise, we find a sample mean of 68.6 hours based on responses from 50 assistant coaches. This figure serves as the center point for constructing our confidence interval. While the sample mean is an estimate, it helps us understand whether or not it reflects the population mean accurately. Importantly, it is the basis upon which our margin of error is applied, forming the backbone of our statistical inference.

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Most popular questions from this chapter

You are to conduct a sample survey to determine the mean family income in a rural area of central Florida. The question is, how many families should be sampled? In a pilot sample of 10 families, the standard deviation of the sample was \(\$ 500 .\) The sponsor of the survey wants you to use the 95 percent confidence level. The estimate is to be within \(\$ 100\). How many families should be interviewed?

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 95 percent confidence level and state that the estimated proportion must be within 2 percent of the population proportion. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?

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Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the Social Security system. The president wants the estimate to be within . 04 of the true proportion. Assume a 95 percent level of confidence. The president's political advisors estimated the proportion supporting the current policy to be \(.60 .\) a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the proportion supporting current policy?

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