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11.4: One Variance Chi-Squared Test

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    Next, we will learn how to apply the One Variance \(X^2\) Procedure to test a statistical claim about a population variance. Consider the following example.

    For a certain drug, based on standards set by the United States Pharmacopeia (USP) - an official public standards-setting authority for all prescription and over-the-counter medicines and other health care products manufactured or sold in the United States, a standard deviation of capsule weights of less than 2 mg is acceptable. A sample of 10 capsules was taken and the weights are provided below:

    99.8, 100.6, 101.7, 100.8, 99.5, 98.9, 101.7, 100.8, 99.8, 98.2

    At \(1\%\) significance level, test the claim that the standard deviation is \(0.7\).

    Note that the sample variance is \(s^2_i=1.32\).

    Now, let’s identify the statistical claim that needs to be tested:

    “the standard deviation is 0.7”

    The key words “standard deviation” suggest that the claim is about the parameter \(\sigma\), so we can symbolically express the claim as

    \(\sigma=0.7\)

    which is equivalent to the claim that the variance is \(0.49\) that is

    \(\sigma^2=0.7^2=0.49\)

    Since the claim is about the population variance, we will use the One Variance \(X^2\) Procedure. Let’s check if all necessary assumptions are satisfied:

    • The sample is simple random
    • The population is normally distributed

    We will use the following template to perform the hypothesis testing:

    In step 1, we will set up the hypothesis:

    Since the claim is in the form of an equation, therefore it must be set as a null hypothesis. The alternative hypothesis, which is always in the form of an inequality, must be carefully chosen depending on the context. Since there is no any indication of one-sided inequality, we set the alternative hypothesis as \(\sigma^2\neq0.49\) and the test is two-tail, that is:

    \(H_0: \sigma^2=0.49\)

    \(H_a: \sigma^2\neq0.49\)

    TT

    In step 2, we will identify the significance level:

    The significance level can always be found in the text of the problem. In our case it is \(1\%\), thus:

    \(\alpha=0.01\)

    In step 3, we will find the test statistic using the formula:

    \(\chi_0^2=\frac{s^2_i(n-1)}{\sigma^2_0}=\frac{1.32\cdot9}{0.49}=24.24\)

    In step 4, we will perform either the critical value approach or p-value approach to test the claim:

    • In critical value approach, we construct the rejection region using the \(X^2\)-curve with \(df=n-1=10-1=9\):

    RR: less than \(\chi^2_{0.995}=1.73\)

    or greater than \(\chi^2_{0.005}=23.59\)

    • In p-value approach, we compute the p-value using the \(X^2\)-curve with \(df=n-1=10-1=9\):

    P-Value:  \(2P(X^2>24.24)=0.0078\)

    In step 5, we will draw the conclusion:

    • In the critical value approach, we must check whether the test statistic is in the rejection region or not. Our test statistic is \(24.24 \) and it is to the right of the critical value \(23.59\), thus it is in the rejection region.
    • In the p-value approach, we must check whether the p-value is less than the significance level or not. Our p-value is \(0.0078\) and it is less than \(\alpha\).

    Both tests suggest that we DO reject the null hypothesis in favor of the alternative.

    In step 6, we will interpret the results:

    Under \(1\%\)  significance level we DO have sufficient evidence to suggest that the variance is not \(0.49\), and therefore the standard deviation is not \(0.7\).

    We discussed how to apply the One Variance \(X^2\) Procedure to test a statistical claim about a population variance (and standard deviation).


    11.4: One Variance Chi-Squared Test is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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