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Chapter 16: Problem 618

A sample of size 10 produced a variance of \(14 .\) Is this sufficient to reject the null hypothesis that \(\sigma^{2}=6\) when tested using a \(.05\) level of significance? Using a \(.01\) level of significance?

Short Answer

Expert verified
In conclusion, the sample variance of 14 is sufficient to reject the null hypothesis (蟽虏 = 6) at the 0.05 level of significance but is insufficient to reject the null hypothesis at the 0.01 level of significance.

Step by step solution

01

State the Hypotheses

First, let's state the null hypothesis (H鈧�) and the alternative hypothesis (H鈧�). H鈧�: The population variance is equal to 6. This can be written as: H鈧�: 蟽虏 = 6 H鈧�: The population variance is not equal to 6. This can be written as: H鈧�: 蟽虏 鈮� 6
02

Identify the Test Statistic and Distribution

Since we are testing the variance, we will use the chi-square distribution. Our test statistic will be the chi-square test statistic, which can be calculated using the following formula: 围虏 = (n - 1) * s虏 / 蟽鈧�虏 where n is the sample size, s虏 is the sample variance (14 in this case), and 蟽鈧�虏 is the hypothesized population variance (6 in this case).
03

Calculate the Test Statistic

Now, let's use the formula to calculate the test statistic: 围虏 = (10 - 1) * 14 / 6 = 9 * 14 / 6 = 126 / 6 = 21
04

Determine the Critical Values

Next, we'll find the critical values for the given levels of significance using a chi-square table or a calculator. Since we have a two-tailed test, we'll need both the lower and upper critical values for each significance level. For a 0.05 level of significance and a sample size of 10 (degrees of freedom = n - 1 = 9): Lower critical value: 围虏(0.025, 9) = 2.70 Upper critical value: 围虏(0.975, 9) = 19.02 For a 0.01 level of significance and a sample size of 10 (degrees of freedom = n - 1 = 9): Lower critical value: 围虏(0.005, 9) = 1.24 Upper critical value: 围虏(0.995, 9) = 23.59
05

Compare the Test Statistic and Decide Whether to Reject the Null Hypothesis

Now, let's compare the test statistic we calculated with the critical values for each level of significance: For a 0.05 level of significance: Since 21 > 19.02 (the upper critical value), the test statistic falls in the rejection region. Therefore, we reject the null hypothesis (H鈧�) at the 0.05 level of significance. For a 0.01 level of significance: Since 21 < 23.59 (the upper critical value), the test statistic does not fall in the rejection region. Therefore, we fail to reject the null hypothesis (H鈧�) at the 0.01 level of significance. In conclusion, the sample variance of 14 is sufficient to reject the null hypothesis (蟽虏 = 6) at the 0.05 level of significance but is insufficient to reject the null hypothesis at the 0.01 level of significance.

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Key Concepts

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

Hypothesis Testing
When we conduct hypothesis testing, we are essentially trying to decide if there is enough evidence from the sample data to infer that a certain condition holds true for the entire population. This process involves two competing hypotheses: the null hypothesis (ewpage( H_0 ewpage)), which is a statement of no effect or no difference, and the alternative hypothesis (ewpage( H_1 ewpage) or ewpage( H_a ewpage)), which is what we suspect might actually be true.

For instance, if we're testing the variance of a population, the null hypothesis might suggest that the population variance is a particular value (ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_1: ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_1: ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_1: ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( H_0: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( test ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( The method we use to test these hypotheses involves four key steps: stating the hypotheses, choosing the appropriate test statistic and its distribution, calculating the test statistic from the data, and making a decision about the hypotheses using a pre-determined level of significance.
.
Chi-Square Distribution
The chi-square distribution is a fundamental tool used in statistics for hypothesis testing, especially when dealing with categorical data or measures of spread such as the variance. This distribution is derived from the sum of the squares of a number of independent and normally distributed random variables. A key property of the chi-square distribution is that it is always positive and skewed to the right, with its shape depending on the degrees of freedom (df).

The degrees of freedom in a chi-square distribution refers to the number of independent values or quantities that can vary in an analysis without violating any constraints. In the context of variance, the degree of freedom is typically the sample size minus one (ewpage( df = n - 1 ewpage)), which is used to estimate the population variance. As the degrees of freedom increases, the chi-square distribution approaches a normal distribution.
Test Statistic
To assess the evidence against the null hypothesis in hypothesis testing, we calculate a test statistic. This statistic distills the sample information into a single number that measures how far our sample statistic, such as the sample mean or variance, is from the parameter specified in the null hypothesis under the assumption that the null is true.

The formula used for the chi-square test for variance is given by: ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( / ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( test statistic ewpage( ewpage( ewpage( / ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( test statistic ewpage( ewpage( = ewpage( ewpage( test statistic ewpage( sample statistic ewpage( ewpage( ewpage( ewpage( ewpage test statistic ewpage( ewpage(ewpage test statistic ewpage( ewpage( ewpage( This statistic is then compared against a critical value from the chi-square distribution to decide whether or not to reject the null hypothesis. The level of significance influences the critical value and hence, the conclusion drawn from the hypothesis test.
Level of Significance
The level of significance, often denoted by ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( significance ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage( ewpage( significance ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage( ewpage(ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( significance ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( ewpage( significance ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( significance ewpage( ewpage(critical values ewpage( significance ewpage( ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage level of significance ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( significance ewpage( ewpage( significance ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( ewpage(ewpage(ewpage(ewpage(ewpage(ewpage level of significance ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage(ewpage( test statistic ewpage( ewpage( ewpage( ewpage( ewpage( ewpage( If the test statistic is more extreme than the critical value, we reject the null hypothesis; if it is less extreme, we do not reject the null hypothesis. It鈥檚 crucial to choose an appropriate level of significance before conducting the test since it directly affects the outcome of the hypothesis test and indicates the degree to which we want to be protected against falsely rejecting a true null hypothesis, known as a Type I error.

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Most popular questions from this chapter

According to a schedule, a given operation is supposed to be performed in \(6.4\) minutes. A study is to be performed to determine whether a particular worker conforms to this standard or whether his performance deviates from this standard, either higher or lower. If the deviation is as much as \(0.5\) minute this should be detected \(95 \%\) of the time. Experience indicates the standard deviation of the distribution of such operations is about \(0.4\) minute, although this figure is not precisely known. At a level of significance of \(.01\), a sample size of how many operations would be needed to perform the study?

The standard deviation of a particular dimension of a metal component is small enough so that it is satisfactory in subsequent assembly. A new supplier of metal plate is under consideration and will be preferred if the standard deviation of his product is not larger, because the cost of his product is lower than that of the present supplier. A sample of 100 items from each supplier is obtained. The results are as follows: New supplier: \(\mathrm{S}_{1}^{2}=.0058\) Old supplier: \(\mathrm{S}_{2}{ }^{2}=.0041\) Should the new supplier's metal plates be purchased? Test at the \(5 \%\) level of significance.

A man has just purchased a trick die which was advertised as not yielding the proper proportion of sixes. He wonders whether the advertising was correct, and tests the advertising claim by rolling the die 100 times. The 100 rolls yielded ten sixes. Should he conclude that the advertising was legitimate?

In one income group, \(45 \%\) of a random sample of people express approval of a product. In another income group, \(55 \%\) of a random sample of people express approval. The standard errors for these percentages are \(.04\) and \(.03\) respectively. Test at the \(10 \%\) level of significance the hypothesis that the percentage of people in the second income group expressing approval of the product exceed that for the first income group.

In a binomial experiment \(\mathrm{H}_{0}\) is that the probability of success on a single trial, \(\mathrm{p}=1 / 3\). Calculate the power of a binomial test with \(\alpha=.05\) and \(\mathrm{n}=10\) when \(\mathrm{H}_{1}\) is \(\mathrm{p}=1 / 2\) and when \(\mathrm{H}_{1}\) is \(\mathrm{p}=2 / 3\). Do the same for \(\mathrm{n}=20\).
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