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Chapter 9: Problem 8

\(z\) test statistic To test \(\mathrm{H}_{0}: p=0.50\) that a population proportion equals 0.50 , the test statistic is a \(z\) -score that measures the number of standard errors between the sample proportion and the \(\mathrm{H}_{0}\) value of \(0.50 .\) If \(z=3.6,\) do the data support the null hypothesis, or do they give strong evidence against it? Explain.

Short Answer

Expert verified
The data give strong evidence against the null hypothesis, as a z-score of 3.6 suggests the sample proportion is significantly different from 0.50.

Step by step solution

01

Understand the Hypotheses

The null hypothesis ( H_0) states that the population proportion is 0.50, while the alternative hypothesis ( H_a) suggests that the population proportion is not equal to 0.50.
02

Interpret the Z-Score

A z-score of 3.6 indicates that the sample proportion is 3.6 standard errors away from the null hypothesis value of 0.50. This z-score value suggests how unusual the sample result is under the assumption that the null hypothesis is true.
03

Determine the Significance Level

Common significance levels are 0.05, 0.01, or 0.001. For this exercise, we need to determine how likely it is to obtain a z-score of 3.6 or more extreme under the null hypothesis. Usually, a z-score beyond approximately 卤1.96 (at 伪 = 0.05) is considered significant.
04

Use Z-Table or P-value Calculator

A z-score of 3.6 corresponds to a very small p-value. We would look at a z-table or use a calculator to find that the p-value associated with a z-score of 3.6 is less than 0.001, indicating very low probability if the null hypothesis is true.
05

Draw Conclusion

Given that the p-value is far smaller than common significance levels (such as 0.05), we reject the null hypothesis. This provides strong evidence against H_0, suggesting the sample proportion is statistically different from 0.50.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), is a statement we aim to test. It typically reflects a position of no change or no difference. For instance, it might suggest that a population parameter鈥攍ike a proportion, mean, or variance鈥攊s equal to a certain value.
In the context of our exercise, the null hypothesis is that the population proportion \( p \) equals 0.50. This means we start with the assumption that there is no deviation from this proportion in the population.
  • The null hypothesis formulates a baseline, allowing us to measure how well the sample data fits under this assumption.
  • Rejecting \( H_0 \) implies that the sample data is inconsistent with it, pointing towards the alternate hypothesis as more plausible.
  • A common misconception is that rejecting \( H_0 \) confirms the alternative hypothesis, whereas it only suggests stronger evidence for it.
Z-Score
The z-score is a statistical measure that indicates how many standard deviations an element is from the mean. In hypothesis testing, the z-score helps us understand how much a sample statistic deviates from the assumed population parameter in the null hypothesis.
For our problem, we calculated a z-score of 3.6. This means that the sample proportion is 3.6 standard errors away from the hypothesized population proportion of 0.50.
  • A z-score can be positive or negative, indicating whether the sample proportion is above or below the hypothesized value.
  • The further the z-score is from zero, the less likely the sample data is to occur if the null hypothesis is true.
  • A high absolute z-score often leads us towards rejecting the null hypothesis in favor of the alternative.
Significance Level
The significance level, denoted as \( \alpha \), is the threshold we set to decide whether to reject the null hypothesis. It's often expressed as a probability and can be thought of as a measure of how much risk of error we are willing to accept.
Common significance levels are 0.05, 0.01, or 0.001, which translate to a 5%, 1%, or 0.1% chance of rejecting the null hypothesis when it is actually true (Type I error).
  • A lower significance level means a stricter criterion for rejecting \( H_0 \), demanding stronger evidence against it.
  • We compare our p-value from the test to \( \alpha \) to make our final decision about \( H_0 \).
  • If the p-value is less than \( \alpha \), we reject the null hypothesis, suggesting the sample provides enough evidence to support the alternative hypothesis.
P-Value
A p-value is a probability that measures the strength of evidence against the null hypothesis. It tells us how likely or unlikely our sample results are if the null hypothesis is true.
In this exercise, the z-score of 3.6 gives us a p-value of less than 0.001. This low p-value means our sample result is rare under the assumption of the null hypothesis.
  • The smaller the p-value, the stronger the evidence against \( H_0 \).
  • P-values help us make informed decisions鈥攊f a p-value is less than the significance level \( \alpha \), we typically reject \( H_0 \).
  • In practice, the p-value allows us to quantify the evidence against the null hypothesis, backing our conclusions with numbers rather than mere speculation.

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Most popular questions from this chapter

A recent study \(^{4}\) considered whether dogs could be trained to detect whether a person has lung cancer or breast cancer by smelling the subject's breath. The researchers trained five ordinary household dogs to distinguish, by scent alone, exhaled breath samples of 55 lung and 31 breast cancer patients from those of 83 healthy controls. A dog gave a correct indication of a cancer sample by sitting in front of that sample when it was randomly placed among four control samples. Once trained, the dogs' ability to distinguish cancer patients from controls was tested using breath samples from subjects not previously encountered by the dogs. (The researchers blinded both dog handlers and experimental observers to the identity of breath samples.) Let \(p\) denote the probability a dog correctly detects a cancer sample placed among five samples when the other four are controls. a. Set up the null hypothesis that the dog's predictions correspond to random guessing. b. Set up the alternative hypothesis to test whether the probability of a correct selection differs from random guessing. c. Set up the alternative hypothesis to test whether the probability of a correct selection is greater than with random guessing. d. In one test with 83 Stage I lung cancer samples, the dogs correctly identified the cancer sample 81 times. The test statistic for the alternative hypothesis in part \(\mathrm{c}\) was \(z=17.7 .\) Report the \(\mathrm{P}\) -value to three decimal places and interpret. (The success of dogs in this study made researchers wonder whether dogs can detect cancer at an earlier stage than conventional methods such as MRI scans.

Too little or too much wine? Wine-pouring vending machines, previously available in Europe and international airports, have become popular in the past few years in the United States. They are even approved to dispense wine in some Walmart stores. The available pouring options are a 5 -ounce glass, a 2.5 -ounce half-glass, and a 1-ounce taste. When the machine is in statistical control (see Exercise 7.43 ), the amount dispensed for a full glass is 5.1 ounces. Four observations are taken each day to plot a daily mean over time on a control chart to check for irregularities. The most recent day's observations were \(5.05,5.15,4.95,\) and \(5.11 .\) Could the difference between the sample mean and the target value be due to random variation, or can you conclude that the true mean is now different from \(5.1 ?\) Answer by showing the five steps of a significance test, making a decision using a 0.05 significance level.

When the 583 female workers in the 2012 GSS were asked how many hours they worked in the previous week, the mean was 37.0 hours, with a standard deviation of 15.1 hours. Does this suggest that the population mean work week for females is significantly different from 40 hours? Answer by: a. Identifying the relevant variable and parameter. b. Stating null and alternative hypotheses. c. Reporting and interpreting the P-value for the test statistic value. d. Explaining how to make a decision for the significance level of 0.01

Example \(4,\) on whether dogs can detect bladder cancer by selecting the correct urine specimen (out of seven), used the normal sampling distribution to find the P-value. The normal distribution P-value approximates a P-value using the binomial distribution. That binomial P-value is more appropriate when either expected count is less than \(15 .\) In Example \(4, n\) was \(54,\) and 22 of the 54 selections were correct. a. If \(\mathrm{H}_{0}: p=1 / 7\) is true, \(X=\) number of correct selections has the binomial distribution with \(n=54\) and \(p=1 / 7\). Why? b. For \(\mathrm{H}_{a}: p>1 / 7,\) with \(x=22,\) the small sample \(\mathrm{P}\) -value using the binomial is \(\mathrm{P}(22)+\mathrm{P}(23)+\cdots+\mathrm{P}(54)\), where \(\mathrm{P}(x)\) denotes the binomial probability of outcome \(x\) with \(p=1 / 7\). (This equals \(0.0000019 .\) ) Why would the P-value be this sum rather than just \(\mathrm{P}(22) ?\)

Results of \(99 \%\) confidence intervals are consistent with results of two- sided tests with which significance level? Explain the connection.
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