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A random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

Short Answer

Expert verified
Once step 2 is done, we'll have observed value and critical value(s). Comparison of them will give us the answer. If observed value is less than or beyond critical value (based on type of test), reject the null hypothesis, otherwise accept it. Answer will depend on outcome after step 2 is done.

Step by step solution

01

Setup Hypotheses

For both parts a) and b), the null and alternative hypotheses have been given in the problem. For instance, for a) hypotheses are \(H_{0}: p=.63\) versus \(H_{1}: p<.63\)
02

Calculate the Test Statistic

The test statistic for a sample proportion is a z-score, which can be calculated using the formula: \[Z = \frac{(p - P_{0})*sqrt(n)}{sqrt(P_{0}*(1-P_{0}))} \]where p is the observed sample proportion, P_{0} is the hypothesized population proportion under the null hypothesis, and n is the sample size. For applied part a), for instance, \(Z_{observed} = \frac{(.60 - .63)*sqrt(200)}{sqrt(.63*(1-.63))}\). You can find the z-score by plugging the values into the formula and solve. For the critical value \(Z_{critical}\), look up z value corresponding to significance level \(\alpha = .01\) in a standard normal probability table. since it's a left-tailed test, z critical will be negative.
03

Compare the Test to the critical value

If the observed z-score is less than the critical z score, we will reject the null hypothesis. So compare and determine the result. Same steps to be followed for part b). For part b) since the test is two-tailed, you will have to find two critical values (at α/2 and -α/2) and if observed value lies beyond any critical value, reject the null hypothesis

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Key Concepts

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

Critical Value
In hypothesis testing, the critical value is a threshold that determines if we reject or fail to reject the null hypothesis. It's directly related to the significance level, \(\alpha\), which represents the probability of making a Type I error or rejecting a true null hypothesis.
To find the critical value for a test with a given significance level, you typically consult a standard normal distribution table (Z-table). This table provides the Z-score that corresponds to your chosen significance level. For a left-tailed test at \(\alpha=0.01\), the critical value is negative since it represents the lower tail of the distribution.
In the two-tailed test context, such as in scenario b) where \(H_1: p eq 0.63\), you split the significance level between two tails, meaning you calculate critical values for \(\pm \alpha/2\). You'll reject the null hypothesis if the observed test statistic falls beyond these critical values, indicating that the sample data is significantly different from the presumed population parameter.
Z-Score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. In hypothesis testing, it transforms the sample proportion into a standard score considering both the sample size and the variability under the null hypothesis.
To calculate the z-score, we use the formula: \[Z = \frac{(p - P_{0})*\sqrt{n}}{\sqrt{P_{0}*(1-P_{0})}} \]where:
  • \(p\) is the observed sample proportion,
  • \(P_{0}\) is the hypothesized population proportion,
  • \(n\) is the sample size.
The z-score indicates how many standard deviations the observed sample proportion is from the hypothesized proportion under the null hypothesis. In a practical sense:
  • If the z-score falls within the critical value(s), we do not reject the null hypothesis.
  • If it falls outside, we reject the null hypothesis, suggesting the sample provides sufficient evidence against the null hypothesis.
Sample Proportion
In statistics, the sample proportion is a crucial concept when estimating the proportion of a characteristic in a population from a sample. It is simply the number of favorable cases divided by the total sample size.
Using the exercise as an example, the sample proportion is \(0.60\) since 120 observations of the 200 met the criteria being evaluated. The sample proportion offers a point estimate of the population proportion, giving you a good idea of what the corresponding population proportion might be.
However, since the sample proportion is based on a subset of the population, it's important to recognize the role of variability and the potential for sampling error. This is accounted for when calculating the standard error of the proportion and applying the z-score formula in hypothesis testing.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\), is a statement that suggests a potential outcome that contradicts the null hypothesis. It reflects what we might suspect to be true or are trying to prove.
In the case of statistical tests, the alternative hypothesis is key because it defines the region of rejection. For example:
  • For part (a) of the exercise, \(H_1: p < .63\), we have a one-tailed test that looks for the sample proportion being significantly less than \(0.63\). The critical region is thus in the left tail of the distribution.
  • For part (b), where \(H_1: p eq 0.63\), it is a two-tailed test, meaning we are interested in deviations in both directions from \(0.63\).
Selecting an alternative hypothesis is not arbitrary; it stems from theoretical prediction, prior research, or specific inquiry goals. Effectively, it guides the interpretation of statistical test results and helps decide whether to reject the null hypothesis.

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Most popular questions from this chapter

A statistician performs the test \(H_{0}: \mu=15\) versus \(H_{1}: \mu \neq 15\) and finds the \(p\) -value to be \(.4546\). a. The statistician performing the test does not tell you the value of the sample mean and the value of the test statistic. Despite this, you have enough information to determine the pair of \(p\) -values associated with the following alternative hypotheses. i. \(H_{1}: \mu<15\) ii. \(H_{1}: \mu>15\) Note that you will need more information to determine which \(p\) -value goes with which alternative. Determine the pair of \(p\) -values. Here the value of the sample mean is the same in both cases. b. Suppose the statistician tells you that the value of the test statistic is negative. Match the \(p\) -values with the alternative hypotheses. Note that the result for one of the two alternatives implies that the sample mean is not on the same side of \(\mu=15\) as the rejection region. Although we have not discussed this scenario in the book, it is important to recognize that there are many real-world scenarios in which this type of situation does occur. For example, suppose the EPA is to test whether or not a company is exceeding a specific pollution level. If the average discharge level obtained from the sample falls below the threshold (mentioned in the null hypothesis), then there would be no need to perform the hypothesis test.

Alpha Airlines claims that only \(15 \%\) of its flights arrive more than 10 minutes late. Let \(p\) be the proportion of all of Alpha's flights that arrive more than 10 minutes late. Consider the hypothesis test $$ H_{0}: p \leq .15 \text { versus } H_{1}: p>.15 $$ Suppose we take a random sample of 50 flights by Alpha Airlines and agree to reject \(H_{0}\) if 9 or more of them arrive late. Find the significance level for this test.

Mong Corporation makes auto batteries. The company claims that \(80 \%\) of its LL.70 batteries are good for 70 months or longer. A consumer agency wanted to check if this claim is true. The agency took a random sample of 40 such batteries and found that \(75 \%\) of them were good for 70 months or longer. a. Using the \(1 \%\) significance level, can you conclude that the company's claim is false? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain.

You read an article that states " 50 hypothesis tests of \(H_{0}: \mu=35\) versus \(H_{1}: \mu \neq 35\) were performed using \(\alpha=.05\) on 50 different samples taken from the same population with a mean of \(35 .\) Of these, 47 tests failed to reject the null hypothesis." Explain why this type of result is not surprising.

Consider the null hypothesis \(H_{0}: \mu=100\). Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100\) b. \(H_{1}: \mu>100\) c. \(H_{1}: \mu<100\)

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