In a random sample of 60 dog owners enrolled in obedience training, it was determined that the mean amountof money spent per owner was $109.33 per class. Assuming the population standard deviation of the amountspent per owner is $12, construct and interpret a 95% confidence interval for the mean amount spent per ownerfor an obedience class.
A) $( \$ 106.29 , \$ 112.37 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.29$ and $\$ 112.37$.
B) $( \$ 106.78 , \$ 111.88 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.78$ and $\$ 111.88$.
C) (\$106.23, \$112.43); we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.23$ and $\$ 112.43$.
D) $( \$ 106.74 , \$ 111.92 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.74$ and $\$ 111.92$.

Question

In a random sample of 60 dog owners enrolled in obedience training, it was determined that the mean amountof money spent per owner was $109.33 per class. Assuming the population standard deviation of the amountspent per owner is $12, construct and interpret a 95% confidence interval for the mean amount spent per ownerfor an obedience class. 
A) $( \$ 106.29 , \$ 112.37 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.29$ and $\$ 112.37$.
B) $( \$ 106.78 , \$ 111.88 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.78$ and $\$ 111.88$.
C) (\$106.23, \$112.43); we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.23$ and $\$ 112.43$.
D) $( \$ 106.74 , \$ 111.92 )$; we are $95 \%$ confident that the mean amount spent per dog owner for a single obedience class is between $\$ 106.74$ and $\$ 111.92$.

Quizplus · Accepted Answer

To construct a 95% confidence interval for the mean, we use the formula: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 109.33$, $\sigma = 12$, $n = 60$), we calculate the margin of error and the confidence interval, which matches the interval given in option A.

In a Random Sample of 60 Dog Owners Enrolled in Obedience