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Chapter 9: Problem 131

Find the test statistic for the hypothesis test: a. \(H_{o}: \sigma^{2}=532\) versus \(H_{a}: \sigma^{2}>532\) using sample information \(n=18\) and \(s^{2}=785\) b. \(H_{o}: \sigma^{2}=52\) versus \(H_{a}: \sigma^{2} \neq 52\) using sample information \(n=41\) and \(s^{2}=78.2\)

Short Answer

Expert verified
The test statistics for the given hypothesis tests, assuming calculations are made correctly, will be the respective results from step 1 and step 2.

Step by step solution

01

Calculate Test Statistic for First Scenario

First, we'll solve for scenario a. Applying the formula for the Chi-Square Test Statistic, we get: \(X^{2}_{a} = \frac{(18-1)785}{532} \)
02

Calculate Test Statistic for Second Scenario

Next, we'll solve for scenario b. Applying the same formula, we get: \(X^{2}_{b} = \frac{(41-1)78.2}{52} \)
03

Simplification of Calculations

Simplify the calculations to find the test statistic for scenarios a and b

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Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used in hypothesis testing. It determines whether there is a significant difference between the expected and observed variances. In the context of testing for variance, this test helps to see if the sample variance significantly deviates from the population variance.
The key formula for the Chi-Square Test Statistic is: \[ \chi^2 = \frac{(n-1) s^2}{\sigma^2} \] This formula involves:
  • \( \chi^2 \): The Chi-Square test statistic
  • \( n \): Sample size
  • \( s^2 \): Sample variance
  • \( \sigma^2 \): Population variance
In the exercise, this formula was applied to test whether the sample variance is greater or not equal to the population variance. Understanding the Chi-Square Test helps you determine if sample data significantly differs from what you'd expect.
Variance
Variance is a measure of how much the values in a data set differ from the mean. It tells us how much variability there is. A high variance signifies that the numbers are spread out quite far from the mean. A low variance indicates that the numbers are closer to the mean.
Formula-wise, variance is the square of the standard deviation.
  • For a population: \( \sigma^2 = \frac{\sum (X - \mu)^2}{N} \)
  • For a sample: \( s^2 = \frac{\sum (X - \bar{X})^2}{n-1} \)
The exercise involved comparing the sample variance \(s^2\) with hypothesized population variance \(\sigma^2\). The purpose was to test if the variability in the sample aligns with what is expected in the population.
Sample Size
Sample size, denoted as \( n \), is the number of observations or data points collected in a sample from a population. It is a crucial factor in hypothesis testing and statistical analysis because:
  • Larger sample sizes generally lead to more reliable results.
  • In tests like the Chi-Square, sample size impacts the test statistic, which is why it's included in the formula.
In the exercise, the sample sizes were 18 and 41. These numbers affect the calculation of the Chi-Square statistic and ultimately the conclusion of the test. Ensuring you have an adequate sample size is essential for drawing meaningful conclusions.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis.
For the Chi-Square Test of variance, the test statistic is calculated as follows: \[ X^2 = \frac{(n-1)s^2}{\sigma^2}\]
  • This formula creates a ratio comparing sample variance and assumed population variance.
  • The degrees of freedom for this test are \( n-1 \), which reflects the sample size adjusted for calculation.
After calculating the test statistic \( X^2 \), it is compared to a critical value from the Chi-Square distribution table or used to compute a p-value. This helps decide if you reject or fail to reject the null hypothesis. Understanding how the test statistic forms the basis for decision-making is fundamental to hypothesis testing.

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Most popular questions from this chapter

You are testing the hypothesis \(p=0.7\) and have decided to reject this hypothesis if after 15 trials you observe 14 or more successes. a. If the null hypothesis is true and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. b. Find the significance level of your test. c. If the true probability of success is \(1 / 2\) and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. d. Calculate the \(p\) -value for your hypothesis test after 13 successes are observed.

You hurry to the local emergency department in hopes of immediate, urgent care, only to find yourself waiting for what seems like hours. The manager of the large emergency department believes that his new procedures have substantially reduced the wait time for the average urgent care patient. He initiates a study to evaluate the wait time. The records of 18 randomly selected patients seen since the new procedures were put in place are checked, and the time between entering the emergency department and being seen by urgent care personnel was observed. The mean wait time was 17.82 minutes with a standard deviation of 5.68 minutes. Estimate the mean wait time using a \(99 \%\) confidence interval. Assume that wait times are normally distributed.

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 recent survey conducted by Lieberman Research Worldwide and Charles Schwab reported that the "High cost of Living" was the top concern that most surprised young adults as they began life on their own. Twenty-six percent reported "High cost of Living" as their top concern. A disbeliever of this information took his own random sample of 500 young adults starting out on their own in an attempt to show that the true percentage for this top concern is actually higher. a. Find the \(p\) -value if 148 of the young adults surveyed put down "High cost of Living" as their top concern. b. Explain why it is important for the level of significance to be established before the sample results are known.

Find: a. \(\chi^{2}(10,0.01)\) b. \(\chi^{2}(12,0.025)\) c. \(\chi^{2}(10,0.95)\) d. \(\chi^{2}(22,0.995)\)
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