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Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes \(n\) is the large-sample \(z\) test appropriate: a. \(H_{0}: p=.2, n=25\) b. \(H_{0}: p=.6, n=210\) c. \(H_{0}: p=.9, n=100\) d. \(H_{0}: p=.05, n=75\)

Short Answer

Expert verified
The large-sample z test is appropriate for the following null-hypotheses and sample sizes: b. \(H_{0}: p=.6, n=210\) and c. \(H_{0}: p=.9, n=100\).

Step by step solution

01

- Calculate expected number of successes and failures

For each hypothesis and sample size, calculate the expected number of successes \(np\) and the expected number of failures \(n(1-p)\). These need to be at least 10 to use the z test.
02

- Determine if expected number of successes and failures are reached

For a., \(n * p = 25 * .2 = 5\) and \(n * (1-p) = 25 * .8 = 20\). Therefore, the z test is not appropriate. For b., \(n * p = 210 * .6 = 126\) and \(n * (1-p) = 210 * .4 = 84\). Thus, the z test is appropriate here. For c., \(n * p = 100 * .9 = 90\) and \(n * (1-p) = 100 * .1 = 10\). So, the z test is appropriate. For d., \(n * p = 75 * .05 = 3.75\) and \(n * (1-p) = 75 * .95 = 71.25\. Thus, the z test is not appropriate here.
03

- Concluding which hypotheses are suitable for the z test

Conclude which hypotheses can use the z test: Hypotheses b. and c. meet the conditions for the z test, while hypotheses a. and d. do not.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It is a statement that assumes there is no significant effect or difference in a particular situation, effectively positing that any observed variations are due to chance rather than a specific cause. In the context of a large-sample z test, the null hypothesis usually claims that the true population proportion is equal to a specified value, just as seen in the exercise with hypotheses like \(H_{0}: p=.2\).

When conducting a hypothesis test, the goal is to determine whether the observed sample data provides enough evidence to reject the null hypothesis. If we find sufficient evidence against it, we may then consider an alternative hypothesis, which suggests that there is indeed an effect or difference. Understanding and correctly stating the null hypothesis is crucial to any statistical test, including the large-sample z test.
Sample Size
Sample size, denoted by \(n\), plays a pivotal role in the validity of statistical tests like the large-sample z test. It represents the number of observations or measurements being analyzed. For the z test to be appropriate, the sample size should be large enough to approximate a normal distribution of sample proportions. This is critical because the z test relies on the central limit theorem, which states that the distribution of sample means will tend to be normal as the sample size increases.

A commonly accepted rule for the large-sample z test is that both the expected number of successes \(np\) and the expected number of failures \(n(1-p)\) should be at least 10. Let's consider an example from the exercise, looking at the hypothesis \(H_{0}: p=0.6\) with a sample size of \(n=210\), the condition \(np > 10\) and \(n(1-p) > 10\) are met, ensuring a sample size large enough for reliable z test results.
Expected Number of Successes
The expected number of successes in a sample is a critical quantity to evaluate before you can perform a large-sample z test. This number is determined by multiplying the sample size \(n\) by the hypothesized population proportion \(p\) stated in the null hypothesis. As demonstrated in the solution, you will calculate it like so: \(n * p\). This value should be at least 10 for the z test to be considered appropriate, as it ensures there is adequate data to approximate a normal distribution.

In our exercise context, for hypothesis a. where \(p = 0.2\), the expected number of successes is \(5\), which does not meet the criterion, rendering the z test inappropriate. A solid understanding of how to calculate and interpret this figure is essential for accurately applying the large-sample z test.
Expected Number of Failures
The expected number of failures is the complement to the expected number of successes when performing a large-sample z test. It is calculated by multiplying the sample size \(n\) by one minus the hypothesized proportion \(1-p\), which gives us the value of \(n(1-p)\). Similar to the expected number of successes, this number must also be at least 10 to justify using a normal approximation. The rule of thumb for both expected values to exceed 10 helps ensure that the sample data's variability is sufficiently captured by the test.

Looking at the provided solution in the exercise, hypothesis d. with an expected number of failures calculated as \(71.25\) seems ample, but because its expected number of successes is only \(3.75\), it still doesn't satisfy the conditions needed for the large-sample z test. Accounting for both successes and failures is critical in determining test appropriateness.
Hypothesis Testing
Hypothesis testing is a systematic method for making statistical decisions using experimental data. It is essentially a way to use sample data to answer research questions. The process involves proposing two hypotheses - the null hypothesis and the alternative hypothesis - and then using a statistical test, like the large-sample z test, to determine which hypothesis the data supports.

The decision in hypothesis testing is made on the basis of whether the observed sample statistics are likely if the null hypothesis is true. This is where the concept of a p-value comes into play; it measures the probability of observing the test statistics if the null hypothesis were correct. If this probability is sufficiently low (typically below a predetermined threshold of 0.05), we reject the null hypothesis in favor of the alternative. It's a cornerstone of inferential statistics and encompasses establishing parameters, calculating statistics, and making inferences about population characteristics based on sample information.

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Most popular questions from this chapter

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

For which of the following \(P\) -values will the null hypothesis be rejected when performing a test with a significance level of .05: a. .001 d. .047 b. .021 e. .148 c. .078

The poll referenced in the previous exercise (“Military Draft Study," AP- lpsos, June 2005 ) also included the following question: "If the military draft were reinstated, would you favor or oppose drafting women as well as men?" Forty-three percent of the 1000 people responding said that they would favor drafting women if the draft were reinstated. Using a .05 significance level, carry out a test to determine if there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.

A television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(\hat{p}\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(p\) denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(p<.9 .\) The appropriate hypotheses are then \(H_{0}: p=.9\) versus \(H_{a}: p<.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds? Use a significance level of .05 .

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