To test the claim that the standard deviation of all women's hair lengths prior to entering the hair salon is greater than 1.0 inch, we can use the chi-squared test for variance.
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Identify Hypotheses:
- Null Hypothesis \( H_0: \sigma = 1.0 , in \)
- Alternative Hypothesis \( H_a: \sigma > 1.0 , in \)
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Determine the Sample Size and Standard Deviation:
- Sample size \( n = 20 \)
- Sample standard deviation \( s = 1.2 , in \)
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Calculate the Test Statistic: The test statistic for the chi-squared test is calculated using the formula: \[ \chi^2 = \frac{(n - 1)s^2}{\sigma_0^2} \] where \( \sigma_0 \) is the standard deviation under the null hypothesis (1.0 inch). \[ \chi^2 = \frac{(20 - 1)(1.2^2)}{1.0^2} = \frac{19 \cdot 1.44}{1.0} = 27.36 \]
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Critical Value: Since this is a one-tailed test with \( \alpha = 0.01 \) and \( n - 1 = 19 \) degrees of freedom, we need to find the critical value from the chi-squared distribution table for \( df = 19 \) at the 0.01 significance level.
Looking up the critical value \( \chi^2_{0.01, 19} \), we find:
- Critical value = 36.191
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Decision Rule:
- If the calculated \( \chi^2 \) statistic (27.36) is greater than the critical value (36.191), we reject the null hypothesis.
Given our analysis, the critical value we need is:
A) 36.191.