/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Let \(\pi\) denote the proportio... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\pi\) denote the proportion of grocery store customers that use the store's club card. For a large sample \(z\) test of \(H_{0}: \pi=.5\) versus \(H_{a}: \pi>.5\), find the \(P\) -value associated with each of the given values of the test statistic: a. \(1.40\) d. \(2.45\) b. \(0.93\) e. \(-0.17\) c. \(1.96\)

Short Answer

Expert verified
The p-values for the given z-scores are: a) 0.0808, d) 0.0071, b) 0.1762, e) 0.4325, c) 0.0250.

Step by step solution

01

Calculate P-Value for z = 1.40

First, use the standard Normal Distribution table or a calculator with a Normal Distribution function to calculate the area to the left of \(z = 1.40\). This gives us \(0.9192\). Since the question wants the probability that \(Z\) is greater than \(1.40,\) the p-value is \(1 - 0.9192 = 0.0808\).
02

Calculate P-Value for z = 2.45

Using the Normal Distribution function or table again, we find that the area to the left of \(2.45\) is \(0.9929\). Therefore, the p-value for this z-value is \(1 - 0.9929 = 0.0071\).
03

Calculate P-Value for z = 0.93

The area to the left of \(0.93\) is \(0.8238\). Therefore, the p-value for this z-value is \(1 - 0.8238 = 0.1762\).
04

Calculate P-Value for z = -0.17

The area to the left of \(-0.17\) is \(0.4325\). But, as \(Z\) is less than \(0,\) we are already in the right side of the curve. So no need to subtract it from \(1\). Thus the p-value for this z-value is \(0.4325\).
05

Calculate P-Value for z = 1.96

Finally, the area to the left of \(1.96\) is \(0.9750\). Therefore, the p-value for this z-value is \(1 - 0.9750 = 0.0250\).

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Key Concepts

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

Z-Test
The z-test is a statistical method that determines whether there is a significant difference between the means of two groups, or if a sample mean significantly differs from a known population mean. It is a type of hypothesis testing that utilizes the standard normal distribution. In the context of a z-test, the test statistic follows a normal distribution if the sample size is sufficiently large due to the Central Limit Theorem.

The test statistic (z-score) measures how many standard deviations an element is from the mean. To calculate it, we use the formula: \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size.

In the exercise, we are asked to calculate the p-value for different z-scores, which represents the probability of observing a value as extreme as the z-score under the null hypothesis. If the p-value is less than a predetermined threshold (commonly 0.05), we reject the null hypothesis in favor of the alternative hypothesis.
Normal Distribution
The Normal Distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean, with its shape known as a 'bell curve.' Most values cluster around a central region, with values tapering off as they go further away from the mean. This distribution is characterized by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).

The total area under the curve is equal to 1, and it is significant in statistics because numerous real-world phenomena fall into this distribution pattern when sample sizes are large. Moreover, the Central Limit Theorem explains how various sample means will form a normal distribution regardless of the original population's distribution, given a large enough sample size.

In hypothesis testing, the normal distribution is used to determine probabilities related to the test statistic. For instance, in our exercise, we calculate the area to the left of given z-scores to find the p-value, which relies on the symmetric property of the normal distribution.
Hypothesis Testing
In statistics, Hypothesis Testing is a method used to decide whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population. In hypothesis testing, two hypotheses are constructed: the null hypothesis \( (H_0) \) and the alternative hypothesis \( (H_a) \). The null hypothesis typically states that there is no effect or no difference, while the alternative hypothesis represents what the researcher is trying to prove.

The process begins by assuming the null hypothesis is true. Then statistical analysis is conducted to determine the likelihood of the sample data if the null hypothesis were true. This likelihood is quantified by the p-value. If the p-value is below a predefined alpha level (usually 0.05), we reject the null hypothesis in favor of the alternative hypothesis. This doesn’t prove the alternative hypothesis; it only suggests that the data are not consistent with the null hypothesis.

The given exercise demonstrates hypothesis testing, where the null hypothesis is \( H_0: \pi = 0.5 \) and the alternative hypothesis is \( H_a: \pi > 0.5 \). By calculating the p-value for various test statistics, we can determine if the sample data provides sufficient evidence to reject the null hypothesis.
Test Statistic
A Test Statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. Essentially, it measures the degree of agreement between a sample statistic and the parameters of the null hypothesis.

In a z-test, the test statistic is the z-value which tells us how many standard deviations our sample mean is from the population mean under the null hypothesis. This is a crucial step in hypothesis testing as it allows us to standardize different data points, making them comparable. The test statistic is used to compute the p-value, which in turn determines whether the results are statistically significant.

In our exercise, when calculating p-values for various z-scores, we are essentially examining where those z-scores lie on the standard normal distribution and how extreme these values are, given the null hypothesis. The test statistics guide us in making an inference about the population parameter after the appropriate calculations.

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Most popular questions from this chapter

The state of Georgia's HOPE scholarship program guarantees fully paid tuition to Georgia public universities for Georgia high school seniors who have a B average in academic requirements as long as they maintain a B average in college. Of 137 randomly selected students enrolling in the Ivan Allen College at the Georgia Institute of Technology (social science and humanities majors) in 1996 who had a B average going into college, \(53.2 \%\) had a GPA below \(3.0\) at the end of their first year ("Who Loses HOPE? Attrition from Georgia's College Scholarship Program," Southern Economic Journal [1999]: 379-390). Do these data provide convincing evidence that a majority of students at Ivan Allen College who enroll with a HOPE scholarship lose their scholarship?

Speed, size, and strength are thought to be important factors in football performance. The article "Physical and Performance Characteristics of NCAA Division I Football Players" (Research Quarterly for Exercise and Sport \([1990]: 395-401\) ) reported on physical characteristics of Division I starting football players in the 1988 football season. Information for teams ranked in the top 20 was easily obtained, and it was reported that the mean weight of starters on top- 20 teams was \(105 \mathrm{~kg}\). A random sample of 33 starting players (various positions were represented) from Division I teams that were not ranked in the top 20 resulted in a sample mean weight of \(103.3 \mathrm{~kg}\) and a sample standard deviation of \(16.3 \mathrm{~kg}\). Is there sufficient evidence to conclude that the mean weight for nontop- 20 starters is less than 105, the known value for top20 teams?

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 twothirds? Use a significance level of \(.05\).

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