Find the confidence interval specified. Assume that the population is normally distributed.
-A random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Estimate the mean number of cars per household for the
Population of households in this neighborhood. Give the 95% confidence interval.
$\begin{array} { l l l l l l l l l l }
2 & 0 & 1 & 2 & 3 & 2 & 1 & 0 & 1 & 4 \
1 & 3 & 2 & 0 & 1 & 1 & 2 & 3 & 1 & 2 \
1 & 0 & 3 & 1 & 1 & 2 & 2 & 1 & 0 & 2
\end{array}$
A)1.3 to 1.7 cars
B)1.1 to 1.9 cars
C)1 to 2 cars
D)1.5 to 1.9 cars

Question

Find the confidence interval specified. Assume that the population is normally distributed.
-A random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Estimate the mean number of cars per household for the
Population of households in this neighborhood. Give the 95% confidence interval. 
$\begin{array} { l l l l l l l l l l } 
2 & 0 & 1 & 2 & 3 & 2 & 1 & 0 & 1 & 4 \
1 & 3 & 2 & 0 & 1 & 1 & 2 & 3 & 1 & 2 \
1 & 0 & 3 & 1 & 1 & 2 & 2 & 1 & 0 & 2
\end{array}$ 
A)1.3 to 1.7 cars
B)1.1 to 1.9 cars
C)1 to 2 cars
D)1.5 to 1.9 cars

Quizplus · Accepted Answer

To find the 95% confidence interval for the mean number of cars per household, we first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). The sample mean is calculated as the sum of all observations divided by the number of observations ($n=30$). The sample standard deviation is calculated using the formula for s in a sample. Then, we use the formula for the 95% confidence interval for the mean, which is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the t-score from the t-distribution for 95% confidence and 29 degrees of freedom (n-1).Given data: $2, 0, 1, 2, 3, 2, 1, 0, 1, 4, 1, 3, 2, 0, 1, 1, 2, 3, 1, 2, 1, 0, 3, 1, 1, 2, 2, 1, 0, 2$Calculating the mean ($\bar{x}$):$\bar{x} = \frac{2+0+1+2+3+2+1+0+1+4+1+3+2+0+1+1+2+3+1+2+1+0+3+1+1+2+2+1+0+2}{30} = \frac{43}{30} \approx 1.433$Calculating the sample standard deviation (s):$s \approx 1.023$ (Calculation not shown for brevity)Using a t-distribution table or calculator for 29 degrees of freedom and a 95% confidence level, we find $t_{\alpha/2} \approx 2.045$.Calculating the margin of error (ME):$ME = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \approx 2.045 \cdot \frac{1.023}{\sqrt{30}} \approx 0.382$Therefore, the 95% confidence interval for the mean number of cars per household is:$\bar{x} \pm ME \approx 1.433 \pm 0.382$This gives us a confidence interval of approximately $1.051$ to $1.815$, which is closest to the option:B) 1.1 to 1.9 cars

Find the Confidence Interval Specified A)13 to 17 Cars B)1

Find the Confidence Interval Specified A)13 to 17 Cars
B)1