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Aptitude tests should produce scores with a large amount of variation so that an administrator can distinguish between persons with low aptitude and persons with high aptitude. The standard test used by a certain industry has been producing scores with a standard deviation of 10 points. A new test is given to 20 prospective employees and produces a sample standard deviation of 12 points. Are scores from the new test significantly more variable than scores from the standard? Use \(\alpha=.01\)

Short Answer

Expert verified
The new test is not significantly more variable than the standard test.

Step by step solution

01

Define the Hypotheses

To determine if the new test is significantly more variable, we formulate the null and alternative hypotheses: - Null Hypothesis \( H_0 \): \( \sigma = 10 \) (The population standard deviation is not greater than the standard.)- Alternative Hypothesis \( H_a \): \( \sigma > 10 \) (The population standard deviation is greater than the standard.)
02

Identify the Test Statistic

For comparing standard deviations, we use the chi-square test for variance. The test statistic is calculated using the formula:\[\chi^2 = \frac{(n - 1) \times s^2}{\sigma^2}\]where \( n = 20 \), \( s = 12 \) (sample standard deviation), and \( \sigma = 10 \) (standard deviation from the standard test).
03

Calculate the Test Statistic Value

Substitute the given values into the chi-square test statistic formula:\[\chi^2 = \frac{(20 - 1) \times (12)^2}{10^2} = \frac{19 \times 144}{100} = \frac{2736}{100} = 27.36\]
04

Determine the Critical Value

Using a chi-square distribution table, find the critical value for a chi-square test with \( n - 1 = 19 \) degrees of freedom and \( \alpha = 0.01 \) significance level. The critical value is approximately 36.19.
05

Compare Test Statistic to Critical Value

Compare the calculated chi-square statistic (27.36) with the critical chi-square value (36.19). Since 27.36 is less than 36.19, we fail to reject the null hypothesis.
06

State the Conclusion

The conclusion based on the chi-square test is that there is not enough evidence to say that the new test is significantly more variable than the standard test at the 0.01 significance level.

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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 critical statistical method used to determine the variability in a dataset. In hypothesis testing, it helps us understand if a sample comes from a population with a specific variance. It's particularly useful when comparing sample and population variances. In the case of the aptitude test, the Chi-square test checks if the new test's variance significantly differs from the standard test's variance.

The test statistic for the Chi-square test is calculated using the formula:
  • \[ \chi^2 = \frac{(n - 1) \times s^2}{\sigma^2} \]
This formula involves:
  • \( n \): the sample size
  • \( s \): the sample standard deviation
  • \( \sigma \): the standard deviation from the population (or standard test)
By comparing this chi-square statistic with a critical value from the chi-square distribution table, we can decide whether we reject the null hypothesis. If the test statistic is greater than the critical value, it indicates a significant variance difference.
Standard Deviation
Standard deviation is a measure that tells us how much the individual data points deviate from the mean. It reflects the amount of variation or dispersion of a set of values. In the context of our aptitude test scenario, understanding standard deviation helps us interpret how the scores spread out from the average.

For the existing standard test, the standard deviation is 10. This means that scores typically vary by 10 units from the mean score. The new test, however, has a sample standard deviation of 12, indicating a difference in how scores are distributed. While the number itself may not suggest a significant difference, statistical tests like the Chi-square test help us deduce whether this change in standard deviation indicates a wider variation among test-takers under the new aptitude test.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold used in hypothesis testing to determine when to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis.

In the aptitude test problem, a significance level of \( \alpha = 0.01 \) was used. This low value suggests that we want to be very confident (99% confidence level) in our findings before rejecting the null hypothesis that states there is no increase in variance. Setting a significance level helps control the likelihood of incorrect conclusions drawn from the statistical test.
Variance Analysis
Variance analysis is a statistical method used to interpret variations within a dataset. It informs us how much individual data points differ from the mean and from one another. In the aptitude test example, variance analysis aids in assessing whether the new test produces results that widely differ compared to the standard one.

Variance itself is calculated as the average of the squared differences from the mean. It provides insight into the dataset's overall dispersion. By conducting variance analysis using statistical tools like the Chi-square test, you can decide if modifications to testing methods significantly impact the spread and consistency of results.

This analysis is crucial for test administrators. Understanding variance helps in making informed decisions to ensure tests effectively differentiate between varied levels of aptitude.

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Most popular questions from this chapter

Let \(Y_{1}, Y_{2}, \ldots, Y_{20}\) be a random sample of size \(n=20\) from a normal distribution with unknown mean \(\mu\) and known variance \(\sigma^{2}=5 .\) We wish to test \(H_{0}: \mu=7\) versus \(H_{a}: \mu>7\) a. Find the uniformly most powerful test with significance level. \(05 .\) b. For the test in part (a), find the power at each of the following alternative values for \(\mu\) : \(\mu_{\alpha}=7.5,8.0,8.5,\) and9.0. c. Sketch a graph of the power function.

What assumptions are made when a Student's \(t\) test is employed to test a hypothesis involving a population mean?

A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects ten shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the accompanying table. $$\begin{array}{ll} \hline \text { Supplier A } & \text { Supplier B } \\ \hline \bar{y}_{1}=1.89 & \bar{y}_{2}=1.85 \\ s_{1}^{2}=.273 & s_{2}^{2}=.094 \\ n_{1}=10 & n_{2}=10 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.10 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(90 \%\) confidence interval for \(\sigma_{\mathrm{B}}^{2}\) and interpret your results.

Define \(\alpha\) and \(\beta\) for a statistical test of hypotheses.

A check-cashing service found that approximately \(5 \%\) of all checks submitted to the service were bad. After instituting a check-verification system to reduce its losses, the service found that only 45 checks were bad in a random sample of 1124 that were cashed. Does sufficient evidence exist to affirm that the check-verification system reduced the proportion of bad checks? What attained significance level is associated with the test? What would you conclude at the \(\alpha=.01\) level?

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