Question

The soil in a certain area has on average 2.9 mg/mA3 of substance A. If 5 soil tests show a concentration of a substance A of 1.8, 2.3, 2.9, 3.4 and 4.1 mg/m^3 in the soil, construct a 95% confidence interval for ơ2. Assume that the population is normally distributed

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Answer 1

$\bar x$ = (1.8 + 2.3 + 2.9 + 3.4 + 4.1)/5 = 2.9

s² = ((1.8 - 2.9)^2 + (2.3 - 2.9)^2 + (2.9 - 2.9)^2 + (3.4 - 2.9)^2 + (4.1 - 2.9)^2)/4 = 0.815

At 95% confidence interval the critical values are $\dpi{100} \chi^2_{\alpha/2, n - 1}$ = $\dpi{100} \chi^2_{0.025, 4}$ = 11.1433

                                                                              $\dpi{100} \chi^2_{1 - \alpha/2, n - 1}$ = $\dpi{100} \chi^2_{0.975, 4}$ = 0.4844

The 95% confidence interval for $\dpi{100} \sigma^2$ is

(n - 1)s²/$\dpi{100} \chi^2_{\alpha/2, n - 1}$ < $\dpi{100} \sigma^2$ < (n - 1)s²/$\dpi{100} \chi^2_{1 - \alpha/2, n - 1}$

= 4 * 0.815/11.1433 < $\dpi{100} \sigma^2$ < 4 * 0.815/0.4844

= 0.29 < $\dpi{100} \sigma^2$ < 6.73