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Consider the null hypothesis \(H_{0}: p=.65\). Suppose a random sample of 1000 observations is taken to perform this test about the population proportion. Using \(\alpha=.05\), 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

Short Answer

Expert verified
The critical value(s) of \(z\) for \n a) a left-tailed test is -1.645, \n b) a two-tailed test are ±1.96, and \n c) a right-tailed test is 1.645.

Step by step solution

01

Identify the population proportion (p)

The population proportion \(p\) given in the problem is 0.65. Hence \(p = 0.65\)
02

Identify the level of significance (\(\alpha\))

The level of significance for the test, \(\alpha\), is given as 0.05. Therefore \( \alpha = 0.05 \)
03

Determine the standard deviation of the sampling distribution

The standard deviation of the sampling distribution can be calculated using the formula \[ \sigma = \sqrt{ \frac{p(1-p)}{n} }\] where \(n\) is the sample size. Substituting \(p = 0.65\) and \(n = 1000\), the standard deviation \(\sigma = 0.0153\) when rounded to four decimal places.
04

Determine the critical value(s) for different types of tests

a) Left-Tailed Test: The critical value of \(z\) can be found by looking up \(\alpha\) in a standard normal table. As this is a left-tailed test, we will look up the value directly. Hence, \( z = -1.645 \). \n b) Two-Tailed Test: Since we divide \(\alpha\) in half for a two-tailed test, we will look up \(\alpha/2 = 0.025\) in the standard normal table. The absolute value of the critical region for a two-tailed test always has two values, one positive and one negative. Hence, \( z = \pm 1.96 \). \n c) Right-Tailed Test: Since this is a right-tailed test, we need to look up \((1-\alpha) = 0.95\) in the standard normal table. Hence, \( z = 1.645 \)

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Key Concepts

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

Population Proportion
In hypothesis testing, the population proportion represents the fraction or percentage of the population that has a particular characteristic of interest. In the provided exercise, the population proportion is denoted by the symbol \( p \), and it is assigned a value of 0.65. This value signifies that 65% of the population possesses a certain attribute or trait. Understanding the population proportion is crucial because it acts as a benchmark against which sample proportions are compared to determine any significant differences. By knowing the population proportion, we set up our null hypothesis, which is a statement asserting that there is no effect or difference strictly due to random sampling.
Level of Significance
The level of significance, symbolized by \( \alpha \), is a threshold that helps decide whether to reject the null hypothesis. In this exercise, \( \alpha \) is given as 0.05, which translates to a 5% risk of concluding that a difference exists when there is none. This 5% serves as a boundary for error probability. If the computed test statistics fall into a region where the probability is less than or equal to \( \alpha \), we reject the null hypothesis. This level helps control the Type I error rate, meaning it limits the chance of wrongly rejecting a true null hypothesis. Choosing \( \alpha \) impacts the sensitivity of the test
  • Lower \( \alpha \): increases confidence in results but may overlook actual effects.
  • Higher \( \alpha \): increases sensitivity but risks more false positives.
Standard Deviation of Sampling Distribution
The standard deviation of the sampling distribution, often known as the standard error, provides insight into the variability of sample proportions. In hypothesis tests involving population proportions, calculating this standard deviation involves a specific formula that accounts for both the proportion and sample size:
\( \sigma = \sqrt{ \frac{p(1-p)}{n} } \)
Plugging in the values from the exercise, where \( p = 0.65 \) and \( n = 1000 \), results in a standard deviation of approximately 0.0153. This measure helps determine how much the sample proportion might vary from the true population proportion by chance. A smaller standard error indicates that the sample proportion is more likely to be close to the population proportion.
Critical Value
In hypothesis testing, the critical value acts as a demarcation point, which guides decisions regarding the null hypothesis. For a given level of significance, \( \alpha \), you can find the critical value by referring to a standard normal distribution table. Critical values vary depending on the type of test:
  • **Left-Tailed Test**: Here, you look up \( \alpha \) directly, as shown in the exercise where \( z = -1.645 \) for \( \alpha = 0.05 \).
  • **Two-Tailed Test**: Here, you need to split \( \alpha \) in half and identify values on both sides of the distribution. For this exercise, \( z = \pm 1.96 \).
  • **Right-Tailed Test**: This involves looking at \( 1-\alpha \), as seen with \( z = 1.645 \) for \( \alpha = 0.05 \).
By using these critical values, you establish rejection regions that determine if observations fall within unlikely ranges under the null hypothesis.
Z-Test
The Z-test is a statistical method used when the population variance is known and the sample size is large, typically greater than 30. It helps test the hypothesis about population proportions or means. In this exercise, the Z-test plays a pivotal role in examining whether the sample proportion significantly deviates from the specified population proportion. To perform a Z-test:
  • Calculate the Z-score using the formula: \( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \)
  • Compare the Z-score with critical values, which decide if you reject the null hypothesis or not.
The Z-test quickly provides evidence of any discordance between the sample findings and the hypothesized population parameter.

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Most popular questions from this chapter

According to the most recent Bureau of Labor Statistics survey on time use in the United States, the average U.S. man spends \(67.20\) minutes per day eating and drinking. Suppose that a survey of 43 Norwegian men resulted in an average of \(81.10\) minutes per day eating and drinking [Note: This value is consistent with the data in a report by the Organization for Economic Cooperation and Development (Source: http://economix.blogs.nytimes.com/2009/05/05/obesity-and- the-fastness-of-food/)]. Assume that the population standard deviation for all Norwegian men is \(18.30\) minutes. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average daily time spent eating and drinking by all Norwegian men is higher than \(67.20\) minutes. What is your conclusion at \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach. Use \(\alpha=.01\).

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.45, \quad H_{1}: p \neq .45, \quad n=100, \quad \hat{p}=.49, \quad \alpha=.10\) b. \(H_{0}: p=.72, \quad H_{1}: p<.72, \quad n=700, \quad \hat{p}=.64, \quad \alpha=.05\) c. \(H_{0}: p=.30, \quad H_{1}: p>.30, \quad n=200, \quad \hat{p}=.33, \quad \alpha=.01\)

In a 2009 nonscientific poll on the Web site of the Daily Gazette of Schenectady, New York, readers were asked the following question: "Are you less inclined to buy a General Motors or Chrysler vehicle now that they have filed for bankruptcy?" Of the respondents, \(56.1 \%\) answered "Yes" (http://www. dailygazette.com/polls/2009/jun/Bankruptcy/). In a recent survey of 1200 adult Americans who were asked the same question, 615 answered "Yes." Can you reject the null hypothesis at the \(1 \%\) significance level in favor of the alternative that the percentage of all adult Americans who are less inclined to buy a General Motors or Chrysler vehicle since the companies filed for bankruptcy is different from \(56.1 \%\) ? Use both the \(p\) -value and the critical-value approaches.

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 \(p=.14\)

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed, and they vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.035\) inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, \(\mu=36\) inches, against the alternative hypothesis, \(\mu \neq 36\) inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.015\) inches. a. 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 adjust it if he chooses the maximum probability of a Type I error to be \(.02 ?\) What if the maximum probability of a Type I error is .10? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02 .\) Does the machine need to be adjusted? What if \(\alpha=.10\) ?

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