A quality‐control engineer is interested in the variability of caffeine concentration (in mg/mL\text{mg/mL}) in a particular brand of espresso. A simple random sample of n=30n=30 cups is taken, and the sample standard deviation is found to be s=0.35 mg/mLs=0.35\text{ mg/mL}. Assuming the population of concentrations is approximately normal, using a 95%95\% confidence level, determine the degrees of freedom, the critical values χL2\chi_{L}^2 and χR2\chi_{R}^2, and the confidence interval estimate (up to two decimal places) for the population standard deviation σ\sigma.

df=29df=29χL2=16.047\chi_{L}^2=16.047χR2=45.722\chi_{R}^2=45.72295%95\% CI: 0.28 mg/mLσ0.47 mg/mL0.28\text{ mg/mL}

df=29df=29χL2=16.791\chi_{L}^2=16.791χR2=46.979\chi_{R}^2=46.97995%95\% CI: 0.30mg/mLσ0.51 mg/mL0.30\text{mg/mL}

df=29df=29χL2=16.047\chi_{L}^2=16.047χR2=45.722\chi_{R}^2=45.72295%95\% CI: 0.16 mg/mLσ0.63 mg/mL0.16\text{ mg/mL}

df=30df=30χL2=16.791\chi_{L}^2=16.791χR2=46.979\chi_{R}^2=46.97995%95\% CI: 0.28 mg/mLσ0.47 mg/mL0.28\text{ mg/mL}

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Problem 1Multiple Choice

A quality‐control engineer is interested in the variability of caffeine concentration (in $mg/mL$ ) in a particular brand of espresso. A simple random sample of $n equals 30$ cups is taken, and the sample standard deviation is found to be $s equals 0.35 mg slash mL$ . Assuming the population of concentrations is approximately normal, using a $95 percent sign$ confidence level, determine the degrees of freedom, the critical values $χ_{L}^{2}$ and $χ_{R}^{2}$ , and the confidence interval estimate (up to two decimal places) for the population standard deviation $σ$ .

$d f equals 29$

$chi sub L squared equals 16.791$

$chi sub R squared equals 46.979$

$95 percent sign$ CI: $0.30 mg slash mL is less than sigma is less than 0.51 mg slash mL$

$d f equals 29$

$chi sub L squared equals 16.047$

$chi sub R squared equals 45.722$

$95 percent sign$ CI: $0.16 mg slash mL is less than sigma is less than 0.63 mg slash mL$

$d f equals 29$

$chi sub L squared equals 16.047$

$chi sub R squared equals 45.722$

$95 percent sign$ CI: $0.28 mg slash mL is less than sigma is less than 0.47 mg slash mL$

$d f equals 30$

$chi sub L squared equals 16.791$

$chi sub R squared equals 46.979$

$95 percent sign$ CI: $0.28 mg slash mL is less than sigma is less than 0.47 mg slash mL$

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