Question

Use the one-standard-deviation x^{2}-test and the one-standard-deviation x^{2}-interval procedure to conduct the required hypothesis test and obtain t

Confidence intervals
ANSWERED
asked 2021-06-19
Use the one-standard-deviation \(\displaystyle{x}^{{{2}}}\)-test and the one-standard-deviation \(\displaystyle{x}^{{{2}}}\)-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. s=7 and n=26
a. \(\displaystyle{H}_{{{0}}}:\sigma={5},{H}_{{{m}{a}{t}{h}{r}{m}{\left\lbrace{a}\right\rbrace}}}:\sigma{>}{5},\alpha={0.01}\).
b. 98% confidence interval.

Expert Answers (1)

2021-06-20

a. Compute the value of the test statistic:
\(\displaystyle{x}^{{{2}}}={\frac{{{n}-{1}}}{{{\sigma_{{{0}}}^{{{2}}}}{s}^{{{2}}}}}}={\frac{{{26}-{1}}}{{{5}^{{{2}}}}}}\dot{{\lbrace}}{7}^{{{2}}}={49}\)
Determine the critical value(s) using table VII with df=n-1=26-1=25
\(\displaystyle{{x}_{{{0.01}}}^{{{2}}}}={44.314}\)
If the test statistic is in the rejection region, then reject the null hypothesis:
\(\displaystyle{44.314}{<}{49}\Rightarrow{R}{e}{j}{e}{c}{t}{H}_{{{0}}}\)
b. Determine the critical values using table VII with \(df=n-1=26-1=25:\)
\(\displaystyle{{x}_{{{1}-{0.01}}}^{{{2}}}}={{x}_{{{0.90}}}^{{{2}}}}={11.524}\)
The boudaries of the confidence interval are then:
\(\displaystyle\sqrt{{{\frac{{{n}-{1}}}{{{{x}_{{{\frac{{\alpha}}{{{2}}}}}}^{{{2}}}}}}}}}\dot{{\lbrace}}{s}=\sqrt{{{\frac{{{26}-{1}}}{{{11.524}}}}}}\dot{{\lbrace}}{7}\approx{10.31}\)
\(\sqrt{\frac{n-1}{x_{1-\frac{\alpha}{2}}^{2}}}\dot{}s=\sqrt{\frac{26-1}{44.314}}\dot{}7\approx 5.258\)

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