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Chapter 9: Problem 83

Consider \(H_{0}=p=.70\) versus \(H_{1}: p \neq .70\). a. A random sample of 600 observations produced a sample proportion equal to .68. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 600 observations taken from the same population produced a sample proportion equal to. \(76 .\) Using \(\alpha=.01\), would you reject the null hypothesis? Comment on the results of parts a and \(\mathrm{b}\).

Short Answer

Expert verified
For the first sample, the null hypothesis applies since the test statistic falls within the critical zone. However, for the second sample, the null hypothesis is rejected as the test statistic lies beyond the critical zone.

Step by step solution

01

Statistically Formulate the Problem

The null hypothesis is \(H_{0}=p=.70\), and the alternative hypothesis is \(H_{1}: p \neq .70\). The significance level, \(\alpha\), for both tests is .01.
02

Compute the Test Statistic for Sample 1

The formula for the test statistic is \(Z = (\overline{P} - P_{0})/(\sqrt{P_{0}(1 - P_{0})/n})\). Here, \(P_{0} = .70\), \(\overline{P} = .68\), and \(n = 600\). Substituting these values into the formula yields \(Z \approx -2.24\).
03

Decide Based on Sample 1

Using an alpha level of .01 means that if the Z-score is less than -2.58 or greater than 2.58 the null hypothesis can be rejected (in a two-tailed test). Since -2.24 lies within that zone, we fail to reject the null hypothesis for the first sample.
04

Compute the Test Statistic for Sample 2

For the second sample, keeping \(P_{0} = .70\) and \(n = 600\), the sample proportion becomes \(\overline{P} = .76\), and calculating the Z-score with the given formula gets us \(Z \approx 4.61\).
05

Decide Based on Sample 2

As with sample 1, we check if 4.61 lies outside of the -2.58 to 2.58 zone. It does, so we reject the null hypothesis for the second sample.
06

Making Observations

Comparing the results from both samples shows that small differences in sample proportions can lead to different decisions on the null hypothesis, especially when sample sizes are large.

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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 is an initial claim that we assume to be true until we have enough evidence to suggest otherwise. It's often denoted as \( H_{0} \). In the given exercise, the null hypothesis claims that the population proportion \( p \) is 0.70. This means that we start by assuming this proportion is correct and look for evidence that might show it isn't.
  • The null hypothesis typically represents "no effect" or "status quo."
  • Our goal in testing is to determine if the observed sample data provides sufficient evidence to reject this hypothesis.
Sample Proportion
A sample proportion is a statistic representing the fraction of a sample with a particular attribute of interest. It's symbolized by \( \overline{P} \). In the context of the exercise, it means the proportion of the 600 observations that reflects the trait we are studying.
  • In part (a), \( \overline{P} = 0.68 \). This is the fraction of the sample sharing the attribute, calculated as 68%.
  • In part (b), \( \overline{P} = 0.76 \), meaning 76% of that sample had the attribute.
  • The sample proportion varies with different samples, influencing the test outcome.
Significance Level
The significance level, denoted by \( \alpha \), is the threshold at which we decide whether an observed effect is statistically significant. It is typically set before testing and represents the probability of rejecting the null hypothesis when it is actually true. In the exercise, this level is 0.01.
  • A low \( \alpha \), like 0.01, indicates a stringent requirement for evidence to reject \( H_{0} \).
  • It implies we're allowing only a 1% chance of incorrectly claiming evidence against \( H_{0} \).
  • This level directly affects the critical values used to evaluate the test statistic.
Test Statistic
The test statistic helps us decide whether to reject the null hypothesis. Calculated from sample data, it measures how far the sample proportion deviates from the null hypothesis proportion, standardized by variability. In this exercise, the Z-test is used for its calculation, given by the formula: \[ Z = (\overline{P} - P_{0})/\sqrt{P_{0}(1 - P_{0})/n} \]
  • Sample 1's test statistic, \( Z \approx -2.24 \), doesn't exceed the critical threshold, so we don't reject \( H_{0} \).
  • Sample 2, however, with \( Z \approx 4.61 \), greatly surpasses our critical values, providing evidence to reject \( H_{0} \).
  • The critical values for a significance level of 0.01 in a two-tailed test are typically \( -2.58 \) and \( 2.58 \).

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Most popular questions from this chapter

Consider \(H_{0}=\mu=20\) versus \(H_{1}: \mu<20 .\) A. What type of error would you make if the null hypothesis is actually false and you fail to reject it? b. What type of error would you make if the null hypothesis is actually true and you reject it?

A random sample of 80 observations produced a sample mean of \(86.50 .\) Find the critical and observed values of \(z\) for each of the following tests of hypothesis using \(\alpha=.10 .\) The population standard deviation is known to be \(7.20\). a. \(H_{0}: \mu=91 \quad\) versus \(\quad H_{1}: \mu \neq 91\) b. \(H_{0}=\mu=91\) versus \(\quad H_{1}: \mu<91\)

Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=23, \quad H_{1}: \mu \neq 23, \quad n=50, \quad \bar{x}=21.25, \quad \sigma=5\) b. \(H_{0}: \mu=15, \quad H_{1}: \mu<15, \quad n=80, \quad \bar{x}=13.25, \quad \sigma=5.5\) c. \(H_{0}: \mu=38, \quad H_{1}: \mu>38, \quad n=35, \quad \bar{x}=40.25, \quad \sigma=7.2\)

The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is \(43.05\) months. The lives of all such batteries have a normal distribution with the population standard deviation of \(4.5\) months. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

Brooklyn Corporation manufactures DVDs. The machine that is used to make these DVDs is known. to produce not more than \(5 \%\) defective DVDs. The quality control inspector selects a sample of \(200 \mathrm{DVDs}\) each week and inspects them for being good or defective. Using the sample proportion, the quality con trol inspector tests the null hypothesis \(p \leq .05\) against the alternative hypothesis \(p>.05\), where \(p\) is th proportion of DVDs that are defective. She always uses a \(2.5 \%\) significance level. If the null hypothesi: is rejected, the production process is stopped to make any necessary adjustments. A recent sample of 200 DVDs contained 17 defective DVD: Using a \(2.5 \%\) significance level, would you conclude that the prod" should be stoppe to make necessary adjustments b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in.par Comment on the results of parts a and \(b\)
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