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When np or nq is not 5 or more, the binomial table (Table \(B\) in Appendix \(A\) ) must be used to find critical values in hypothesis tests involving proportions. Show that \(z=\frac{\hat{p}-p}{\sqrt{p q / n}}\) can be derived from \(z=\frac{X-\mu}{\sigma}\) by substituting \(\mu=n p\) and \(\sigma=\sqrt{n p q}\) and dividing both numerator and denominator by \(n .\)

Short Answer

Expert verified
The formula \( z=\frac{\hat{p}-p}{\sqrt{p q / n}} \) for proportions is derived by expressing \( X \) as \( \hat{p}n \) and simplifying.

Step by step solution

01

Understanding the Problem

The aim is to show how the formula for the test statistic involving proportions, \( z=\frac{\hat{p}-p}{\sqrt{p q / n}} \), can be derived from the normal distribution's formula for the standard score \( z=\frac{X-\mu}{\sigma} \). We will substitute the parameters with those related to the binomial distribution.
02

Identify Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( np \) and the standard deviation \( \sigma \) by \( \sqrt{npq} \), where \( p \) is the probability of success, \( q=1-p \) is the probability of failure, and \( n \) is the number of trials.
03

Substitute in the Formula

Substituting the binomial mean and standard deviation into \( z=\frac{X-\mu}{\sigma} \), we get:\[ z = \frac{X-np}{\sqrt{npq}} \]
04

Express Proportions

To work with proportions, express \(X\) as \(\hat{p}n\), where \(\hat{p}\) is the sample proportion. Thus, substituting \(X = \hat{p}n\) into the equation gives us:\[ z = \frac{\hat{p}n - np}{\sqrt{npq}} \]
05

Factor Out \(n\)

Extract the factor \(n\) from the numerator:\[ z = \frac{n(\hat{p} - p)}{\sqrt{npq}} \]
06

Divide Numerator and Denominator by \(n\)

Divide both the numerator and the denominator by \(n\) to get:\[ z = \frac{\hat{p} - p}{\sqrt{pq/n}} \]This shows that the test statistic for proportions can be derived from the generic formula for the standard score.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states. It is commonly used in statistics when dealing with experiments or tests that have a fixed number of trials, two possible outcomes (success or failure), a constant probability of success in each trial, and independent trials. If any one of these conditions changes, the distribution might not fit well.

For example, flipping a coin a specific number of times or checking if a product passes a quality test are scenarios where the binomial distribution applies. The parameters for a binomial distribution are:
  • \( n \): Number of trials.
  • \( p \): Probability of success in a single trial.
  • \( q = 1-p \): Probability of failure in a single trial.
The mean of the binomial distribution, \( \mu \), is calculated as \( np \). The variance is \( npq \), from which the standard deviation is derived as \( \sigma = \sqrt{npq} \).

Understanding these parameters is crucial for analyzing data that follows a binomial distribution, particularly in hypothesis testing scenarios.
Test Statistic
In statistics, a test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis. The test statistic can follow different probability distributions depending on the nature of the data and the type of test being conducted.

For example, when testing proportions based on binomially distributed data, the calculated test statistic is often a z-score. This is because the sampling distribution of the sample proportion is approximately normal under certain conditions, thanks to the central limit theorem.

The formula for the test statistic involving proportions is given by \( z = \frac{\hat{p} - p}{\sqrt{pq/n}} \), where:
  • \( \hat{p} \): Sample proportion.
  • \( p \): Population proportion.
  • \( n \): Sample size.
This test statistic is crucial for evaluating hypotheses about proportions, as it allows comparison against a standard normal distribution to see if the observed sample proportion is significantly different from the hypothesized population proportion.
Normal Distribution
The normal distribution is a continuous probability distribution characterized by its bell-shaped curve. It is symmetric about the mean, implying that the data near the mean are more frequent in occurrence than data far from the mean.

This distribution is defined by two parameters: the mean \( \mu \) and the standard deviation \( \sigma \). When data follows a normal distribution, about 68% of it falls within one standard deviation of the mean, 95% within two, and 99.7% within three, which is often referred to as the empirical rule.

In hypothesis testing, the z-score is used to measure the distance between an observation and the population mean in units of standard deviation. For data relying on the binomial distribution, as the sample size increases, the distribution of the sample mean tends to become normal (a property known as the central limit theorem). Thus, the normal distribution allows us to use z-scores effectively even in a binomial context, under the condition that both \( np \) and \( nq \) are equal to or greater than 5.
Proportions
A proportion is simply a part of a whole expressed in the context of a total. In statistics, it usually refers to the fraction of the sample or population that has a certain attribute. For instance, if 30 out of 100 surveyed people prefer chocolate ice cream, the proportion preferring chocolate is 0.3.

In hypothesis testing, we often compare a sample proportion (\( \hat{p} \)) to a hypothesized population proportion (\( p \)). The formula \( \frac{\hat{p} - p}{\sqrt{pq/n}} \) helps in determining whether there is enough statistical evidence to conclude that the sample comes from a population with a different proportion.

Here, the sample proportion \( \hat{p} \) provides an estimate of the population proportion. The role of the test statistic in this context is to standardize the comparison by accounting for the expected variability due to sampling. Proportions are crucial in various fields, including biology, business, and social sciences, as they provide insights into the make-up of populations and are central to many statistical analyses.

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Most popular questions from this chapter

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Soft Drink Consumption A researcher claims that the yearly consumption of soft drinks per person is 52 gallons. In a sample of 50 randomly selected people, the mean of the yearly consumption was 56.3 gallons. The standard deviation of the population is 3.5 gallons. Find the \(P\) -value for the test. On the basis of the \(P\) -value, is the researcher's claim valid?

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Stopping Distances A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet. A group of automotive engineers decided to conduct a study of its school buses and found that for 20 randomly selected buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet. The standard deviation of the population was 3 feet. Test the claim that the average stopping distance of the company's buses is actually less than 264 feet. Find the \(P\) -value. On the basis of the \(P\) -value, should the null hypothesis be rejected at \(\alpha=0.01 ?\) Assume that the variable is normally distributed.

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Dress Shirts In a previous study conducted several years ago, a man owned on average 15 dress shirts. The standard deviation of the population is \(3 .\) A researcher wishes to see if that average has changed. He selected a random sample of 42 men and found that the average number of dress shirts that they owned was \(13.8 .\) At \(\alpha=0.05,\) is there enough evidence to support the claim that the average has changed?

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. $$ \begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array} $$

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Sleep Time A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.2 hours. The researcher feels that college students do not sleep 7.2 hours on average. The researcher randomly selected 15 students and found that on average they slept 8.3 hours. The standard deviation of the sample is 1.2 hours. At \(\alpha=0.05,\) is there enough evidence to say that college students do not sleep 7.2 hours on average?

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