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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. \(n=40\) and \(p=.11\) b. \(n=100\) and \(p=.73\) c. \(n=80 \quad\) and \(\quad p=.05\) d. \(n=50\) and \(p=.14\)

Short Answer

Expert verified
Tests of hypothesis about the population proportion using the normal distribution are only valid for situation B and D because in those cases both \(np\) and \(n(1-p)\) are greater than 5.

Step by step solution

01

Situation A Analysis

Let's evaluate the first situation where \(n=40\) and \(p=.11\). We calculate \(np = 40 * 0.11 = 4.4\) and \(n(1 - p) = 40 * (1 - 0.11) = 35.6\). Because \(np < 5\), we cannot use the normal distribution to make a test of hypothesis about the population proportion.
02

Situation B Analysis

Now, let's look at situation B where \(n=100\) and \(p=.73\). We have \(np = 100 * 0.73 = 73\) and \(n(1 - p) = 100 * (1-0.73) = 27\). These both are greater than 5, so we can use the normal distribution to make a test of hypothesis about the population proportion.
03

Situation C Analysis

For situation C where \(n=80\) and \(p=.05\), we get \(np = 80 * 0.05 = 4\) and \(n(1 - p) = 80 * (1-0.05) = 76\). Again, because \(np < 5\), using the normal distribution to make a test of hypothesis about the population proportion is not suitable.
04

Situation D Analysis

In situation D, \(n=50\) and \(p=.14\). Hence, \(np = 50 * 0.14 = 7\) and \(n(1 - p) = 50 * (1-0.14) = 43\). Both are greater than 5, so we can indeed use the normal distribution to make a test of hypothesis about the population proportion.

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Key Concepts

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

Sample Size
When dealing with statistics, the term "sample size" refers to the number of individual data points collected from a population. It's denoted by the symbol \( n \). In hypothesis testing, especially with proportions, the sample size plays a crucial role in determining the reliability of any conclusions drawn. A larger sample size tends to provide a better reflection of the population characteristics.

The sample size influences:
  • The margin of error: Larger sample sizes usually result in a smaller margin of error.
  • The variability of data: A larger \( n \) typically reduces the variability or spread of data points.
  • The applicability of statistical tests: For certain tests, like using the normal distribution to approximate the sampling distribution of a sample proportion, the sample size must meet specific criteria—usually being large enough so that both \( np \) and \( n(1-p) \) are greater than 5.
It’s important to select an adequate sample size to ensure your hypothesis testing results are valid and meaningful.
Population Proportion
Population proportion, symbolized as \( p \), indicates the fraction or percentage of the total population that possesses a particular characteristic. This proportion can be practically anything, such as the proportion of adults who own a smartphone or the proportion of defective items in a batch.

When sampling from a population, the sample proportion (often denoted by \( \hat{p} \)) provides an estimate of the true population proportion. However, due to the variability of sample data, \( \hat{p} \) might differ from \( p \).
  • Estimate reliability: Larger samples have sample proportions that are closer to the true population proportion.
  • Bias and precision: A random sample ensures unbiased estimation of the population proportion.
Understanding the population proportion helps set the groundwork for hypothesis testing, specifically when determining whether a sample provides enough evidence to support a particular hypothesis.
Hypothesis Testing
Hypothesis testing is a core statistical method used to determine whether there is enough evidence in a sample of data to conclude that a certain condition is true for the entire population. In the context of population proportions, hypothesis testing often revolves around deciding if the sample data significantly supports or contradicts a pre-established assumption (null hypothesis) about the population proportion.

The basic steps for hypothesis testing are:
  • Establish a null hypothesis \( H_0 \) and an alternative hypothesis \( H_a \).
  • Define a significance level (often \( \alpha = 0.05 \)).
  • Calculate a test statistic based on the sample data.
  • Make a decision based on the test statistic and significance level—either rejecting or not rejecting the null hypothesis.
Effective hypothesis testing provides a systematic way of evaluating claims about a population proportion using sample data.
np Condition
The \( np \) condition is a vital criterion in determining if we can justifiably employ the normal distribution to approximate the distribution of the sample proportion. Specifically, when performing hypothesis testing on proportions, the sample size \( n \) and the estimated population proportion \( p \) need to satisfy certain conditions.

The conditions are:
  • \( np \geq 5 \)
  • \( n(1-p) \geq 5 \)
These conditions ensure that the sample size is large enough for the sampling distribution of the sample proportion \( \hat{p} \) to be approximately normally distributed. If these are not met, the results of hypothesis testing may not be valid or reliable due to the skewness and increased variability inherent in smaller sample sizes or extreme proportions.

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Most popular questions from this chapter

In a 2005 Energy Information Administration report (http://www.cia.doc.gov/cmeu/reps/enduse er01 us.html), the average U.S. household uses 10,654 kilowatt-hours of electricity per year. A random sample of 85 houses built in the last 12 to 24 months showed that they had an average electricity usage of 10,278 kilowatt-hours per year. Assume that the population standard deviation is 1576 kilowatt-hours per year. a. Using the critical-value approach, can you conclude that the average annual clectricity usage of all houses built in the last 12 to 24 months is less than 10,654 kilowatt-hours? Use \(\alpha=.01\). b. What is the Type I error in part a? Explain. What is the probability of making this error in part a? c. Will your conclusion in part a change if the probability of making a Type I error is zero? d. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.01\) ?

A random sample of 14 observations taken from a population that is normally distributed produced a sample mean of \(212.37\) and a standard deviation of \(16.35 .\) Find the critical and observed values of \(t\) and the ranges for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.10\). a. \(H_{0}: \mu=205\) versus \(H_{1}: \mu \neq 205\) b. \(H_{0}: \mu=205\) versus \(H_{1}: \mu>205\)

A real estate agent claims that the mean living area of all single-family homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? What will your conclusion be if \(\alpha=.01 ?\)

According to the University of Wisconsin Dairy Marketing and Risk Management Program (http:// future.aae.wisc.edu/index.html), the average retail price of 1 gallon of whole milk in the United States for April 2009 was \(\$ 3.084\). A recent random sample of 80 retailers in the United States produced an average milk price of \(\$ 3.022\) per gallon, with a standard deviation of \(\$ .274 .\) Do the data provide significant evidence at the \(1 \%\) level to conclude that the current average price of 1 gallon of milk in the United States is lower than the April 2009 average of \(\$ 3.084\) ?

Consider \(H_{0}: p=.45\) versus \(H_{1}: p<.45\). a. A random sample of 400 observations produced a sample proportion equal to .42. Using \(\alpha=.025\), would you reject the null hypothesis? b. Another random sample of 400 observations taken from the same population produced a sample proportion of .39. Using \(\alpha=.025\), would you reject the null hypothesis?

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