/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 NBA Free Throws In Exercise 6.10... [FREE SOLUTION] | 91Ó°ÊÓ

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NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.

Short Answer

Expert verified
Approximately 92.79% of samples will have a sample proportion greater than 0.80

Step by step solution

01

Understand the problem

Understand that we need to find the percentage of samples where the proportion of successes is greater than 80%, given that the population proportion of success is 75%. The number of trials in each sample is \( n = 200 \). Because of large sample size, the sample proportion follows approximately normal distribution.
02

Compute the mean and standard deviation

Calculate mean and standard deviation of the distribution of sample proportions using the formulas: Mean \(\mu = p\) and Standard Deviation \(\sigma = \sqrt{ P(1-P)/n }\) where P is the success rate in the population and n is the number of trials in each sample. Substituting the given values, \(\mu = 0.75\) and \(\sigma = \sqrt{ 0.75 * 0.25 / 200 } \approx 0.0343\)
03

Compute the Z-score

Find the Z-score for a sample proportion of 80% using the formula: \( Z = (X - \mu) / \sigma \) where X is the sample proportion, \(\mu\) is the population proportion, and \(\sigma\) is the standard deviation. Substituting the calculated and provided values, \( Z = (0.8 - 0.75) / 0.0343 \approx 1.46 \)
04

Find percentage of samples

The Z-score tells us how many standard deviations our value is from the mean. Knowing that, we can use the standard normal distribution table or calculator to find the percentage of values that are higher than our z-score. The corresponding value for Z=1.46 is about 0.0721 or 7.21%. However this is the value less than 80%, hence the value more than 80% would be \(1-0.0721 = 0.9279\) or 92.79%.

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Key Concepts

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

Sample Proportion
In statistical analysis, the sample proportion is a critical metric; it indicates the percentage of times an event occurs within a sample. For example, if we look at NBA free throws, and we're interested in the frequency of successful shots, the sample proportion is the total number of successful free throws divided by the total number of attempts in our sample.

Crucially, the sample proportion can serve as an estimator for the population proportion—the long-term success rate of free throws in the NBA, in this case. However, it's vital to note that sample proportions will vary from sample to sample; this variation gives rise to what we call the 'sampling distribution of the sample proportion'.

When we pick a large enough sample (typically larger than 30), according to the Central Limit Theorem, this sampling distribution can often be approximated by a normal distribution, making it easier to handle statistically. For a sample of size n=200, as in our NBA free throw example, this assumption of normality generally holds true.
Standard Deviation
Standard deviation is an all-important statistical measure that quantifies the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much the values in a dataset deviate from the mean (average) value. In the context of the sample proportion, the standard deviation provides insight into how much the proportions of successful free throws in different samples might vary around the population proportion.

The formula for the standard deviation of a sample proportion is given by \( \sigma = \sqrt{P(1-P)/n} \) where \( P \) is the population proportion of successes and \( n \) is the sample size. As the sample size increases, the standard deviation decreases. This relationship indicates that larger samples tend to produce a sample proportion closer to the population proportion, hence implying less variability or more precise estimations.
Normal Distribution
The concept of normal distribution is one of the fundamental pillars of statistics. Often referred to as the 'bell curve', this continuous probability distribution is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In practice, when we're dealing with a large number of repeated samples, like drawing multiple samples of 200 NBA free throws, the distribution of the sample proportions tends to resemble a normal distribution. This aspect of normal distribution allows us to make further inferences and calculate probabilities using the standard normal distribution, which is a normal distribution with a mean of 0 and a standard deviation of 1. For example, determining the percentage of samples with a sample proportion greater than 80% becomes feasible when we assume a normal distribution.
Z-score
A Z-score is a fundamental statistic that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. It is calculated using the formula: \( Z = (X - \mu) / \sigma \) where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

A positive Z-score indicates that the value is above the mean, while a negative Z-score shows it's below the mean. In our NBA example, calculating the Z-score helps us to understand how unusual a sample proportion of 80% is in comparison to the expected proportion of 75%. With the calculated Z-score, we can refer to Z-tables or use statistical software to find out the probability of observing a sample proportion greater than 80%, giving us key insights into the performance of NBA players' free throws relative to historical data.

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Most popular questions from this chapter

The AllCountries dataset shows that the percent of the population living in rural areas is \(57.3 \%\) in Egypt and \(21.6 \%\) in Jordan. Suppose we take random samples of size 400 people from each country, and compute the difference in sample proportions \(\hat{p}_{E}-\hat{p}_{J}\) where \(\hat{p}_{E}\) represents the sample proportion living in rural areas in Egypt and \(\hat{p}_{J}\) represents the sample proportion living in rural areas in Jordan. (a) Find the mean and standard deviation of the distribution of differences in sample proportions, \(\hat{p}_{E}-\hat{p}_{J}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. (c) Using the graph drawn in part (b), are we likely to see a difference in sample proportions as large in magnitude as \(0.4 ?\) As large as \(0.5 ?\) Explain.

Exercise B.5 on page 305 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in mg) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8\). Figure 6.26 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise B.5, we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

Exercise 4.86 on page 263 introduces a matched pairs study in which 47 participants had cell phones put on their ears and then had their brain glucose metabolism (a measure of brain activity) measured under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). Brain glucose metabolism is measured in \(\mu \mathrm{mol} / 100 \mathrm{~g}\) per minute, and the differences of the metabolism rate in the on condition minus the metabolism rate in the off condition were computed for all participants. The mean of the differences was 2.4 with a standard deviation of \(6.3 .\) Find and interpret a \(95 \%\) confidence interval for the effect size of the cell phone waves on mean brain metabolism rate.

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0\) ) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-sample } \mathrm{T} \text { for HeartRate }\\\ &\begin{array}{lrrrr} \text { Status } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \end{array} \end{aligned} $$ Difference \(=m u(0)-m u(1)\) Estimate for difference: -2.12 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0(\) vs not \(=):\) T-Value \(=-0.45\) P-Value \(=0.653 \quad \mathrm{DF}=60\)

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