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Chapter 9: Problem 76

In each of the following cases, do you think the sample size is large enough to use the normal distribution to make a test of hypothesis about the population proportion? Explain why or why not. a. \(n=30\) and \(p=.65\) b. \(n=70\) and \(p=.05\) c. \(n=60\) and \(p=.06\) d. \(n=900\) and \(p=.17\)

Short Answer

Expert verified
a. Yes, sample size is large enough. b. No, sample size is not large enough. c. No, sample size is not large enough. d. Yes, sample size is large enough.

Step by step solution

01

Identify the condition for normal approximation

The condition for using the normal approximation to the binomial distribution for hypothesis testing is that the sample size should be large enough such that both \(np\) and \(n(1-p)\) are greater than 5.
02

Apply the condition to each case

a. For the case with \(n=30\) and \(p=.65\), calculate \(np=30*.65=19.5\) and \(n(1-p)=30*.35=10.5\), both values are greater than 5, so the sample size is large enough. b. For the case with \(n=70\) and \(p=.05\), calculate \(np=70*.05=3.5\) and \(n(1-p)=70*.95=66.5\), here \(np<5\), so the sample size is not large enough. c. For the case with \(n=60\) and \(p=.06\), calculate \(np=60*.06=3.6\) and \(n(1-p)=60*.94=56.4\), here \(np<5\), so the sample size is not large enough. d. For the case with \(n=900\) and \(p=.17\), calculate \(np=900*.17=153\) and \(n(1-p)=900*.83=747\), both values are greater than 5, so the sample size is large enough.
03

Summarize the results

In summary, for a and d, the sample size are large enough to use the normal distribution for hypothesis testing about population proportion, while for b and c, the sample size are not large enough.

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Key Concepts

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

Understanding Binomial Distribution
The binomial distribution is a fundamental concept in statistics. It describes the number of successes in a fixed number of independent trials, each with the same probability of success.
- Think of it like flipping a coin multiple times.
  • If the coin lands on heads, it's a success. If not, it's a failure.
  • Each trial is independent, meaning the outcome of one trial doesn't affect the others.
  • The probability of success (landing on heads) remains constant.
In the context of hypothesis testing, especially when dealing with proportions, the binomial distribution can help us determine the likelihood of a certain number of successes. However, when we are handling large sample sizes, we often approximate the binomial distribution using the normal distribution. This is useful because normal calculations can be simpler and more familiar to most people.
Exploring Population Proportion
Population proportion is a way to express the fraction or percentage of a population that has a particular attribute.
- Imagine you want to know the percentage of people in a city who prefer tea over coffee.
  • The population proportion, denoted by \(p\), is the ratio of tea over coffee lovers to the total population.
  • For example, if 65 out of 100 people prefer tea, the population proportion is \(p = 0.65\).
Knowing the population proportion helps in estimating various population characteristics, and forms an integral part of hypothesis testing, where we make assumptions about this proportion and test them using sample data.
Introduction to Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. In simple terms, it's about making an educated guess, and then testing if this guess is likely to be true.
- Initially, we set up two competing hypotheses:
  • The null hypothesis \((H_0)\), which assumes no change or effect. For example, the proportion of people who like tea is 0.5.
  • The alternative hypothesis \((H_a)\), which assumes some change or effect. For example, the proportion is not 0.5.
After setting our hypotheses, we utilize sample data to test these claims. Choosing an appropriate distribution (like the normal distribution for large samples from a binomial distribution) is crucial to correctly assessing whether to accept or reject the null hypothesis.
The Importance of Sample Size
Sample size is a critical factor in determining the reliability of hypothesis tests. It influences how well our sample data represents the broader population.
- In the context of using normal approximation for binomial distributions:
  • The rule we follow is checking if both \(np\) and \(n(1-p)\) are greater than 5."np" is the expected number of successes, and "\(n(1-p)\)" is the expected number of failures.
  • This ensures that the sample size accurately reflects the population, allowing us to use the normal distribution as an approximation.
For instance, in the provided exercise, sample sizes and proportions determine whether the approximation is valid. Larger sample sizes tend to yield better approximations and more reliable hypothesis test results.

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Most popular questions from this chapter

What are the five steps of a test of hypothesis using the critical value approach? Explain briefly.

Consider \(H_{0}: \mu=29\) versus \(H_{1}: \mu \neq 29 .\) A random sample of 25 observations taken from this population produced a sample mean of \(25.3 .\) The population is normally distributed with \(\sigma=8 .\) al Calculate the \(p\) -value. b. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.05 ?\) c. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.01\) ?

According to the U.S. Postal Service, the average weight of mail received by Americans in 2011 through the Postal Service was \(57.2\) pounds (The New York Times, December 4,2011 ). One hundred randomly selected Americans were asked to keep all their mail for last year. It was found that they received an average of \(55.3\) pounds of mail last year. Suppose that the population standard deviation is \(8.4\) pounds. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average weight of mail received by all Americans last year was less than \(57.2\) pounds. Will you reject the null hypothesis at \(\alpha=.01 ?\) Explain. What if \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach. Will you reject the null hypothesis at \(\alpha=.01\) ? What if \(\alpha=.025\) ?

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, cal culates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and ad. justed. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) oun Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control. inspector decide to stop the machine and readjust it if she chooses the maximum probability of Type I error to be \(.01\) ? What if the maximum probability of a Type I error is b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01 .\) Does the machine need to be adjusted? What if \(\alpha=05\) ?

A random sample of 80 observations produced a sample mean of \(86.50 .\) Find the critical and observed values of \(z\) for each of the following tests of hypothesis using \(\alpha=.10 .\) The population standard deviation is known to be \(7.20\). a. \(H_{0}: \mu=91 \quad\) versus \(\quad H_{1}: \mu \neq 91\) b. \(H_{0}=\mu=91\) versus \(\quad H_{1}: \mu<91\)
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