Question

A group of social workers conclude that the skyscrapers of New York City, obtain a variance equal to 33.4 in the amount of stories. A sample is taken of 42 New York City Skyscrapers; the sample contained a mean of 62.9 stories with a standard deviation of 16.54. Test the claim of the social workers, in the sense that the sample contains a different variance in stories Null Hypothesis, Ho Research Hypothesis, H Degrees of Freedom Alpha Chi-Square Critical Value Chi Square Test Statistic P-Value Conclusion using Critical Statistic Conclusion using P-Value Conclusion in Context

Answer 1

H₀: $\sigma^2$ = 33.4

H₁: $\sigma^2 \neq$ 33.4

DF = 42 - 1 = 41

Alpha = 0.05

At alpha = 0.05, the critical values are $\chi^2_{0.025, 41}$ = 25.2145

                                                            $\chi^2_{0.975, 41}$ = 60.5606

The test statistic $\chi^2$ = (n - 1)s²/ $\sigma^2$ = 41 * (16.54)^2/33.4 = 335.82

P-value = 2 * P( $\chi^2$ > 335.82)
              = 2 * (1 - P( $\chi^2$ < 335.82)

              = 2 * (1 - 1) = 0

As the test statistic value does not lie between the critical values, so we should reject the null hypothesis.

As the P-value is less than alpha (0 < 0.05) , we should reject the null hypothesis.

So there is sufficient evidence to support the claim that the sample cointains a different variance in stories.