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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 \(p=.7\) c. \(n=80 \quad\) and \(\quad p=.05\) d. \(n=50\) and \(p=.14\)

Short Answer

Expert verified
For \(n=40\) and \(p=.11\) and for \(n=80\) and \(p=.05\) the sample size is not large enough, but for \(n=100\) and \(p=.7\) and for \(n=50\) and \(p=.14\) the sample size is indeed large enough.

Step by step solution

01

Evaluating the First Case

Check if the conditions for \(n=40\) and \(p=.11\) are satisfied. Calculate np and n(1-p) and check whether both are greater than or equal to 5. For np, we have \(40 * .11 = 4.4\) and for n(1-p) we have \(40 * (1-.11) = 35.6\). Neither np nor n(1-p) are both greater than or equal to 5, therefore the sample size is not large enough to justify the use of normal distribution.
02

Evaluating the Second Case

Check if the conditions for \(n=100\) and \(p=.7\) are satisfied. Calculate np and n(1-p) and check whether both are greater than or equal to 5. For np, we have \(100 * .7 = 70\) and for n(1-p) we have \(100 * (1-.7) = 30\). Both np and n(1-p) are greater than or equal to 5, therefore the sample size is large enough to justify the use of normal distribution.
03

Evaluating the Third Case

Check if the conditions for \(n=80\) and \(p=.05\) are satisfied. Calculate np and n(1-p) and check whether both are greater than or equal to 5. For np, we have \(80 * .05 = 4\) and for n(1-p) we have \(80 * (1-.05) = 76\). Neither np nor n(1-p) are both greater than or equal to 5, therefore the sample size is not large enough to justify the use of normal distribution.
04

Evaluating the Fourth Case

Check if the conditions for \(n=50\) and \(p=.14\) are satisfied. Calculate np and n(1-p) and check whether both are greater than or equal to 5. For np, we have \(50 * .14 = 7\) and for n(1-p) we have \(50 * (1-.14) = 43\). Both np and n(1-p) are greater than or equal to 5, therefore the sample size is large enough to justify the use of normal distribution.

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Key Concepts

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

Sample Size
The term _sample size_ refers to the number of observations or data points collected in a sample. It's crucial when conducting statistical tests, such as hypothesis tests about population proportions, because the size can affect the results' accuracy and reliability. When using a normal distribution to approximate binomial distribution, the sample size needs to be large enough.

A common rule of thumb is that both the product of the sample size and sample proportion ( \( np \) ) and the product of the sample size and the compliment ( \( n(1-p) \) ) need to be at least 5. This ensures that the normal approximation of the binomial distribution is accurate.

In situations where the sample size is small, other methods such as exact tests may be preferable.
Population Proportion
Population proportion is a metric that represents the fraction of the population that possesses a particular characteristic. In our scenario, this is denoted by _p_. For instance, if we want to study the proportion of people supporting a certain candidate in an election, _p_ would represent that proportion in the whole population.

When conducting hypothesis testing about population proportions, knowing _p_ is vital. It helps in determining the necessary sample size and in calculating errors in estimation. While the true population proportion might be unknown, estimates are often drawn from sample data.

Population proportions are central to hypothesis testing, as they offer a basis against which sample data is tested.
Hypothesis Testing
Hypothesis testing is a statistical method used to decide whether there is enough evidence to reject a null hypothesis. It's widely used in research, including studies on population proportions. The null hypothesis usually states that there is no effect or no difference, and the alternate hypothesis states the opposite.

When hypothesis testing for population proportions, researchers often use sample data to infer about the population. This involves comparing the sample proportion to the hypothesized population proportion. A key step is determining the _p-value_, which helps decide whether to reject the null hypothesis.

Validity of hypothesis testing often rests on the assumption of a normal distribution, especially for large sample sizes, as it simplifies calculations and interpretation.
NP Condition
The _np condition_ is a critical criterion for determining whether a normal distribution can approximate a binomial distribution. This condition requires that both \( np \) and \( n(1-p) \) be no less than 5.

The purpose of this condition is to ensure that the sample probability distribution closely resembles the normal distribution, making the approximation valid. If either condition fails, the normal approximation might not accurately reflect the underlying binomial distribution, potentially leading to erroneous conclusions.

Therefore, it's essential to check this condition before conducting hypothesis tests involving population proportions to ascertain the results' reliability and validity.

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Most popular questions from this chapter

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 \(\alpha=05\) ?

For each of the following significance levels, what is the probability of making a Type I error? \(\begin{array}{lll}\text { a. } \alpha=.10 & \text { b. } \alpha=.02 & \text { c. } \alpha=.005\end{array}\)

Consider the null hypothesis \(H_{0}: p=.25 .\) Suppose a random sample of 400 observations is taken to perform this test about the population proportion. Using \(\alpha=.01\), show the rejection and nonrejection regions and find the critical value(s) of \(z\) for a a. left-tailed test b. two-tailed test c. right-tailed test

The manager of a service station claims that the mean amount spent on gas by its customers is $$\$ 15.90$$ per visit. You want to test if the mean amount spent on gas at this station is different from $$\$ 15.90$$ per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known. \(9.73\) A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

The police that patrol a heavily traveled highway claim that the average driver exceeds the 65 miles per hour speed limit by more than 10 miles per hour. Seventy-two randomly selected cars were clocked by airplane radar. The average speed was \(77.40\) miles per hour, and the standard deviation of the speeds was \(5.90\) miles per hour. Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.02\) ?

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