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91Ó°ÊÓ

Let \(p\) denote the proportion of grocery store customers who use the store's club card. For a largesample \(z\) test of \(H_{0}: p=.5\) versus \(H_{a}: p>.5,\) 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 as follows: a) 0.0808, b) 0.1783, c) 0.0250, d) 0.0071, e) 0.5675.

Step by step solution

01

Understand the values and tables

Before deriving the \(P\) -value, the Z-score must be understood. The Z-score is a measure of how many standard deviations an element is from the mean. Therefore, the given values (1.40,0.93,1.96,2.45 and -0.17) are all Z-scores. To calculate the \(P\) -value, a Standard Normal Distribution Table, or Z-table, is needed. The Z-table provides the probability that a normally distributed random variable Z is less than or equal to a given value.
02

Finding the p-values (Unilateral test)

Because the alternative hypothesis \(H_{a}\) suggests that \(p>.5\), this is a one-sided test, and the \(P\) -values should be calculated as such. To find the \(P\) -value from the Z-table, one must locate the corresponding value and then subtract from 1, since we are looking for \(p>.5\). For instance, for Z=1.40, the Z-table gives a probability of 0.9192, so \(P=(1-0.9192)=0.0808\). The same strategy is applied to the remaining Z-scores.
03

Interpret the p-values

The \(P\) -value is the probability of observing a result as extreme as the test statistic, assuming the null hypothesis is true. If the \(P\) -value is less than the significance level, which is typically set at 0.05, then the null hypothesis is rejected. In this context, a lower \(P\) -value implies that more customers are likely to use the store's club card.

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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 used to determine whether there is a significant difference between sample and population parameters. It is based on the assumption that the samples are drawn from a normal distribution. In the context of hypothesis testing, a Z-test may be used to ascertain whether an observed sample mean differs significantly from a known population mean, or in the case of proportions, whether the observed proportion differs from a hypothesized value.

The Z-test involves calculating a Z-score, which represents the number of standard deviations a data point is from the population mean. The Z-score is then used to find the P-value from the standard normal distribution to determine if the results are significant.
Standard Normal Distribution
The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is the distribution that Z-scores are based on. In practical terms, the Z-score of any data point tells us how far and in what direction that data point is from the mean, measured in units of standard deviation. When all the Z-scores are plotted, they form the standard normal distribution. This curve is symmetrical, and most values are within three standard deviations from the mean.

Understanding the properties of the standard normal distribution is critical for performing Z-tests, as it serves as the reference for determining the probability (P-value) of observing a particular value of the test statistic under the null hypothesis.
Hypothesis Testing
Hypothesis testing is a method used in statistics to decide between two competing hypotheses about a population parameter, based on sampled data. The first step in hypothesis testing is to establish a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_a\)).

The null hypothesis is typically a statement of no effect or no difference, whereas the alternative hypothesis represents a new claim or the effect we wish to test. Using a sample, we calculate a test statistic (like the Z-score in Z-tests) and utilize this statistic to calculate the P-value, which helps us determine whether to reject the null hypothesis or not.
Null Hypothesis
The null hypothesis (\(H_0\)) is a statement used in statistical tests that assumes there is no significant difference or effect. It represents the 'status quo' and is the hypothesis that researchers seek to test against. In the exercise provided, the null hypothesis states that the proportion of customers who use the store's club card is 0.5, or in other words, half of the customers use the card. The null hypothesis serves as a baseline that the test statistic is compared against to determine the P-value.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) represents a statement that is accepted if the null hypothesis is rejected. It proposes what we suspect might be true instead of the null hypothesis. Unlike the null hypothesis, the alternative hypothesis indicates a new effect or change. In our example, the alternative hypothesis is that more than half (\(p > 0.5\)) of the grocery store customers use the store's club card. It provides direction for the statistical test and influences the type of test performed, such as one-tailed or two-tailed tests.
Significance Level
The significance level, often denoted by \(\alpha\), is a threshold used to determine whether or not to reject the null hypothesis. It is a probability value that defines the unlikely range for the P-value if the null hypothesis is true. Common significance levels are 0.01, 0.05, and 0.10. In most social science research, the significance level is set at 0.05, meaning there is a 5% risk of rejecting the null hypothesis when it is, in fact, true. This level is chosen to balance the risk of making an error with the need for reasonable evidence against the null hypothesis.
Z-table
A Z-table, also known as the standard normal table, is a mathematical table that allows us to find the probability that a standard normal variable falls within a specified range. The Z-table displays the cumulative probabilities of the standard normal distribution. As Z-scores follow the standard normal distribution, the table can be used in conducting Z-tests.

When you're doing a hypothesis test, you use the Z-table to find the P-value by locating the area to the left of the Z-score. If the test is one-tailed (as in our exercise, where we're interested in values greater than a certain point), you would subtract this area from 1 to get the P-value. For the given values in the exercise, you would find the corresponding cumulative probabilities for each Z-score and perform the appropriate calculations to determine the P-values.

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Most popular questions from this chapter

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 Economist collects data each year on the price of a Big Mac in various countries around the world. The price of a Big Mac for a sample of McDonald's restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llllll}3.80 & 5.89 & 4.92 & 3.88 & 2.65 & 5.57\end{array}\) \(\begin{array}{ll}6.39 & 3.24\end{array}\) The mean price of a Big Mac in the U.S. in May 2009 was \(\$ 3.57\). For purposes of this exercise, assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using \(\alpha=.05\).

In \(2006,\) Boston Scientific sought approval for a new heart stent (a medical device used to open clogged arteries) called the Liberte. This stent was being proposed as an alternative to a stent called the Express that was already on the market. The following excerpt is from an article that appeared in The Wall Street Journal (August 14,2008 ): Boston Scientific wasn't required to prove that the Liberte was 'superior' than a previous treatment, the agency decided - only that it wasn't "inferior' to Express. Boston Scientific proposed - and the FDA okayed - a benchmark in which Liberte could be up to three percentage points worse than Express meaning that if \(6 \%\) of Express patients' arteries reclog, Boston Scientific would have to prove that Liberte's rate of reclogging was less than \(9 \%\). Anything more would be considered 'inferior.'... In the end, after nine months, the Atlas study found that 85 of the patients suffered reclogging. In comparison, historical data on 991 patients implanted with the Express stent show a \(7 \%\) rate. Boston \(S\) cientific then had to answer this question: Could the study have gotten such results if the Liberte were truly inferior to Express?" Assume a \(7 \%\) reclogging rate for the Express stent. Explain why it would be appropriate for Boston Scientific to carry out a hypothesis test using the following hypotheses: \(H_{0}: p=.10\) \(H_{a}: p<.10\) where \(p\) is the proportion of patients receiving Liberte stents that suffer reclogging. Be sure to address both the choice of the hypothesized value and the form of the alternative hypothesis in your explanation.

Let \(\mu\) denote the mean diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(\quad n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(\quad n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(\quad n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(\quad n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) g. Is the way in which \(\beta\) changes as \(n, \alpha, \sigma,\) and \(\mu\) vary consistent with your intuition? Explain.

Suppose that you are an inspector for the Fish and Game Department and that you are given the task of determining whether to prohibit fishing along part of the Oregon coast. You will close an area to fishing if it is determined that fish in that region have an unacceptably high mercury content. a. Assuming that a mercury concentration of \(5 \mathrm{ppm}\) is considered the maximum safe concentration, which of the following pairs of hypotheses would you test: $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu>5 $$ or $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu<5 $$ Give the reasons for your choice. b. Would you prefer a significance level of .1 or .01 for your test? Explain.

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