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Chapter 9: Problem 84

Consider \(H_{0}: p=.45\) versus \(H_{1}: p<.45\). a. A random sample of 400 observations produced a sample proportion equal to .42. Using \(\alpha=.025\), would you reject the null hypothesis? b. Another random sample of 400 observations taken from the same population produced a sample proportion of .39. Using \(\alpha=.025\), would you reject the null hypothesis?

Short Answer

Expert verified
The null hypothesis should be rejected or failed to be rejected based on the comparison of calculated z-scores with the critical z-score for each of the two samples.

Step by step solution

01

Calculate z-score for the first sample

The formula for the z-score is: \(Z = \frac{(P - p)}{\sqrt{\frac{pq}{n}}}\), where \(P\) is the sample proportion, \(p\) is the population proportion under the null hypothesis, \(q = 1-p\) and \(n\) is the sample size. So for the first sample, we substitute these variables with their respective values: \(P = 0.42\), \(p = 0.45\), \(q = 0.55\), \(n = 400\). Let's calculate the z-score.
02

Compare the calculated z-score with the critical z-score

Once you've calculated the z-score, it needs to be compared with the critical z-score for \(\alpha = 0.025\) in the negative direction. The critical z-score for \(\alpha = 0.025\) is -1.96 (from the standard normal distribution table). If the calculated z-score is less than -1.96, the null hypothesis should be rejected.
03

Calculate z-score for the second sample

Repeat the process for the second sample using its sample proportion \(P = 0.39\) and all the other respective variables.
04

Compare the calculated z-score with the critical z-score again

Compare the calculated z-score for the second sample with the critical z-score of -1.96 and make a conclusion about the null hypothesis.

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Key Concepts

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

z-score calculation
The z-score is a statistical measure that helps determine how far away a sample proportion is from the population proportion under the null hypothesis. It assists in deciding whether the observed sample result is unusual enough to reject the null hypothesis. To calculate the z-score for a sample, you use the formula:
  • \( Z = \frac{(P - p)}{\sqrt{\frac{pq}{n}}} \)
  • Where:
    • \( P \) is the sample proportion, which is the proportion observed in the sample.
    • \( p \) is the hypothesized population proportion under the null hypothesis.
    • \( q \) which equals \( 1 - p \) is the other proportion that complements \( p \).
    • \( n \) is the size of the sample, or number of observations.
First, you plug the sample proportion, the hypothesized population proportion, and the sample size into the formula to find the z-score. A calculated z-score can tell you how many standard deviations the sample proportion is from the hypothesized proportion. This is crucial for hypothesis testing.
critical z-score
A critical z-score is a key value that separates the rejection region from the non-rejection region in hypothesis testing. For a given level of significance, denoted by \( \alpha \), the critical z-score marks the threshold in the standard normal distribution. To decide if there is enough evidence to reject the null hypothesis, compare the calculated z-score to the critical z-score. If the calculated z-score is more extreme, or falls beyond the critical value in the tail of the distribution, then the null hypothesis can be rejected.
  • In this example, the critical z-score for a significance level of \( \alpha = 0.025 \) in the negative direction is \( -1.96 \).
  • This value indicates the number of standard deviations below the mean where the rejection region begins.
This threshold helps in accepting or rejecting the null hypothesis based on the z-score calculated from the sample data.
null hypothesis
The null hypothesis is a starting point in hypothesis testing that assumes no effect or no difference from the status quo. In the context of this problem, the null hypothesis is denoted as \( H_0: p = 0.45 \). Here, \( p \) represents the population proportion we are testing against.The goal of hypothesis testing is to determine whether there is enough statistical evidence in the sample data to reject the null hypothesis in favor of the alternative hypothesis \( H_1 \).
  • The null hypothesis asserts that any observed difference is due to sampling variability rather than an actual effect or difference in the population.
  • It's crucial to have a clear null hypothesis to measure the strength of evidence provided by sample data.
If the test results in rejecting the null hypothesis, it suggests that the sample provides enough evidence that the actual population proportion might be different than 0.45.
sample proportion
The sample proportion is an important element in hypothesis testing as it represents the fraction of subjects in the sample with a particular characteristic. It serves as an estimate of the corresponding population proportion.
  • In this context, sample proportion \( P \) is computed from the sample data and is used in the calculation of the z-score.
  • For example, if you have a sample of 400 observations, and 42% of these share a certain characteristic, the sample proportion is \( P = 0.42 \).
    • Similarly, with a sample proportion of 0.39 for another sample, it implies 39% of the observations meet a certain criterion.
The computed sample proportions are compared against the population proportion under the null hypothesis in statistical tests to determine if there are significant differences between observed samples and what is expected under the null hypothesis. This comparison is what helps in making data-driven decisions.

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Most popular questions from this chapter

A soft-drink manufacturer claims that its 12 -ounce cans do not contain, on average, more than 30 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of \(5 \%\). Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value and \(\alpha=.05\) ?

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.

For each of the following examples of tests of hypothesis about the population proportion, show the rejection and nonrejection regions on the graph of the sampling distribution of the sample proportion. a. A two-tailed test with \(\alpha=.10\) b. A left-tailed test with \(\alpha=.01\) c. A right-tailed test with \(\alpha=.05\)

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

Perform the following tests of hypotheses for data coming from a normal distribution. a. \(H_{0}: \mu=94.80, H_{1}: \mu<94.80, n=12, \quad \bar{x}=92.87, s=5.34, \quad \alpha=.10\) b. \(H_{0}: \mu=18.70, \quad H_{1}: \mu \neq 18.70, n=25, \quad \bar{x}=20.05, s=2.99, \quad \alpha=.05\) c. \(H_{0}: \mu=59, \quad H_{1}: \mu>59, \quad n=7, \quad \bar{x}=59.42, s=.418, \quad \alpha=.01\)
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