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91Ó°ÊÓ

Calculate the test statistic \(z \star\) used in testing the following: a. \(H_{o}: p=0.70\) vs. \(H_{a}: p>0.70,\) with the sample \(n=300\) and \(x=224\) b. \(H_{o}: p=0.50\) vs. \(H_{a}: p<0.50,\) with the sample \(n=450\) and \(x=207\) c. \(H_{o}: p=0.35\) vs. \(H_{a}: p \neq 0.35,\) with the sample \(n=280\) and \(x=94\) d. \(H_{o}: p=0.90\) vs. \(H_{a}: p>0.90,\) with the sample \(n=550\) and \(x=508\)

Short Answer

Expert verified
The test statistics for the given sets are: a. \(z^* = 1.67\), b. \(z^* = -2.89\), c. \(z^* = -0.72\), and d. \(z^* = 1.92\).

Step by step solution

01

Calculate for Set a.

For set a, we have \(H_0: p = 0.70\), \(n = 300\), and \(x = 224\). First, calculate \(\hat{p} = x/n = 224/300 = 0.7467\). Then, plug these values into the formula to get \(z^* = (0.7467 - 0.70)/\sqrt{(0.70(1-0.70)/300)} = 1.67 .
02

Calculate for Set b.

For set b, we have \(H_0: p = 0.50\), \(n = 450\), and \(x = 207\). Again, first calculate \(\hat{p} = x/n = 207/450 = 0.46\). Then, substitute into the formula to get \(z^* = (0.46 - 0.50) \ / \ \sqrt{(0.50(1-0.50) \ / \ 450) } = -2.89 .
03

Calculate for Set c.

For set c, we have \(H_0: p = 0.35\), \(n = 280\), and \(x = 94\). First, calculate \(\hat{p} = x/n = 94/280 = 0.3357\). Then substitute these values into the formula to get \(z^* = (0.3357 - 0.35) \ / \ \sqrt{(0.35(1-0.35) \ / \ 280) } = -0.72 .
04

Calculate for Set d.

For set d, we have \(H_0: p = 0.90\), \(n = 550\), and \(x =508\). Start by calculating \(\hat{p} = x/n = 508/550 = 0.9236\). Then use the formula to calculate \(z^* = (0.9236 - 0.90) \ / \ \sqrt{(0.90(1-0.90) \ / \ 550) } = 1.92 .

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Key Concepts

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

Z-Test Calculation
When conducting hypothesis tests involving proportions, the z-test is a practical way to determine if there's a significant difference between the observed sample proportion and the hypothesized population proportion. This test uses a z-score to assess how far away the sample proportion (\(\hat{p}\)) is from the population proportion (\(p\)) specified in the null hypothesis. The formula for the z-test calculation for proportions is as follows:\[z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\]Where:
  • \(\hat{p}\) is the sample proportion.
  • \(p\) is the population proportion under the null hypothesis.
  • \(n\) is the sample size.
In practice, you first calculate the sample proportion and then use this formula to compute the z-score. This score tells you how many standard deviations away from the mean your sample proportion is. Depending on the significance level, you can determine if the observed data is sufficiently different from the hypothesis to reject it.
Sample Proportion
The sample proportion is a crucial part of hypothesis testing when comparing population proportions. It represents the fraction of the sample that has the characteristic of interest. In mathematical terms, it's denoted by \(\hat{p}\) and is calculated simply as:\[\hat{p} = \frac{x}{n}\]Where:
  • \(x\) is the number of successes or favorable outcomes in the sample.
  • \(n\) is the total number of observations or the sample size.
The sample proportion provides the observed proportion which you compare against the population proportion in your hypothesis. It's a snapshot of your sample's characteristics and is used to conduct further calculations, like the z-test, to validate your hypotheses.
Null and Alternative Hypotheses
In hypothesis testing, null and alternative hypotheses form the foundation for statistical experiments. They are two competing statements that your data will help confirm or refute.The null hypothesis, denoted by \(H_{0}\), is a statement that asserts there is no effect or difference. It's the hypothesis we traditionally aim to test against and, ultimately, reject if enough evidence suggests otherwise.For example:
  • For set a, \(H_{0}: p = 0.70\).
  • For set b, \(H_{0}: p = 0.50\).
The alternative hypothesis, represented by \(H_{a}\), proposes what we might conclude if the null hypothesis is rejected. It's often consistent with the research question you're investigating:
  • For set a, \(H_{a}: p > 0.70\).
  • For set b, \(H_{a}: p < 0.50\).
Understanding these concepts ensures that hypotheses are tested correctly and results are interpreted accurately.
Standard Error
The standard error is a key concept when performing z-test calculations in hypothesis testing. It quantifies the amount of variation or dispersion of the sample proportion from the true population proportion.For proportions, the standard error (\(SE\)) is calculated using the formula:\[SE = \sqrt{\frac{p(1-p)}{n}}\]Where:
  • \(p\) is the population proportion under the null hypothesis.
  • \(n\) is the sample size.
The standard error decreases as the sample size increases, making the estimate of the population proportion more precise. It essentially determines the width of the margin of error in your estimate.In hypothesis testing, the standard error helps standardize the difference between the sample proportion and the hypothesized proportion, allowing you to derive the z-score and decide if the observed difference is statistically significant. By understanding and applying the standard error correctly, you can ensure that your hypothesis tests are robust.

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Most popular questions from this chapter

In obtaining the sample size to estimate a proportion, the formula \(n=[z(\alpha / 2)]^{2} p q / E^{2}\) is used. If a reasonable estimate of \(p\) is not available, it is suggested that \(p=0.5\) be used because this will give the maximum value for \(n\). Calculate the value of \(p q=p(1-p)\) for \(p=0.1,0.2,0.3, \ldots, 0.8,0.9\) in order to obtain some idea about the behavior of the quantity \(p q\).

It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

A production process is considered out of control if the produced parts have a mean length different from \(27.5 \mathrm{mm}\) or a standard deviation that is greater than \(0.5 \mathrm{mm} .\) A sample of 30 parts yields a sample mean of \(27.63 \mathrm{mm}\) and a sample standard deviation of \(0.87 \mathrm{mm} .\) If we assume part length is a normally distributed variable, does this sample indicate that the process should be adjusted to correct the standard deviation of the product? Use \(\alpha=0.05\).

You are interested in comparing the null hypothesis \(p=0.8\) against the alternative hypothesis \(p<0.8 .\) In 100 trials you observe 73 successes. Calculate the \(p\) -value associated with this result.

Julia Jackson operates a franchised restaurant that specializes in soft ice cream cones and sundaes. Recently she received a letter from corporate headquarters warning her that her shop is in danger of losing its franchise because the average sales per customer have dropped "substantially below the average for the rest of the corporation." The statement may be true, but Julia is convinced that such a statement is completely invalid to justify threatening a closing. The variation in sales at her restaurant is bound to be larger than most, primarily because she serves more children, elderly, and single adults rather than large families who run up big bills at the other restaurants. Therefore, her average ticket is likely to be smaller and exhibit greater variability. To prove her point, Julia obtained the sales records from the whole company and found that the standard deviation was \(2.45\)dollar per sales ticket. She then conducted a study of the last 71 sales tickets at her store and found a standard deviation of \(2.95\)dollar per ticket. Is the variability in sales at Julia's franchise, at the 0.05 level of significance, greater than the variability for the company?

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