A sample standard deviation and sample size are given. Use the one-standard-deviation x 2 -test to conduct the requiredhypothesis test.
-$\mathrm { s } = 13.1 , \mathrm { n } = 22$
$\mathrm { H } _ { 0 } : \sigma = 16.1 , \mathrm { H } _ { \mathrm { a } } : \sigma

Question

A sample standard deviation and sample size are given. Use the one-standard-deviation x 2 -test to conduct the requiredhypothesis test.
-$\mathrm { s } = 13.1 , \mathrm { n } = 22$
$\mathrm { H } _ { 0 } : \sigma = 16.1 , \mathrm { H } _ { \mathrm { a } } : \sigma < 16.1 , \alpha = 0.01$  
A)Test statistic: $\mathrm { X } ^ { 2 } = 13.903$. Critical value $= 8.897$. Do not reject $\mathrm { H } _ { 0 }$.
B) Test statistic: $\mathrm { X } ^ { 2 } = 13.903$. Critical value $= 8.897$. Reject $\mathrm { H } _ { 0 }$.
C) Test statistic: $X ^ { 2 } = 13.903$. Critical value $= 38.932$. Reject $\mathrm { H } _ { 0 }$.
D) Test statistic: $\mathrm { X } ^ { 2 } = 13.903$. Critical value $= 38.932$. Do not reject $\mathrm { H } _ { 0 }$.

Quizplus · Accepted Answer

The test statistic is calculated as $\mathrm { X } ^ { 2 } = ( \mathrm { n } - 1 ) \mathrm { s } ^ { 2 } / \sigma ^ { 2 }$. Plugging in the given values, we get $\mathrm { X } ^ { 2 } = ( 22 - 1 ) 13.1 ^ { 2 } / 16.1 ^ { 2 } = 13.903$. The critical value for a one-tailed test with $\alpha = 0.01$ and $\mathrm { n } - 1 = 21$ degrees of freedom is 8.897. Since the test statistic is greater than the critical value, we reject the null hypothesis and conclude that $\sigma < 16.1$.

A Sample Standard Deviation and Sample Size Are Given s=13.1,n=22\mathrm { s } = 13.1 , \mathrm { n } = 22s=13.1,n=22

A Sample Standard Deviation and Sample Size Are Given $\mathrm { s } = 13.1 , \mathrm { n } = 22$