A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normally distributed. $\$ 3.60 \quad \$ 4.50 \quad \$ 2.80 \quad \$ 6.30 \quad \$ 2.60 \quad \$ 5.20 \quad \$ 6.75 \quad \$ 4.25 \quad \$ 8.00 \quad \$ 3.00$ Find the 95% confidence interval for the true mean.
A)($3.39, $6.01)
B)($2.11, $5.34)
C)($4.81, $6.31)
D)($1.35, $2.85)

Question

Quizplus · Accepted Answer

To find the 95% confidence interval for the true mean of normally distributed incomes, we first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). The sample mean is $\frac{3.60 + 4.50 + 2.80 + 6.30 + 2.60 + 5.20 + 6.75 + 4.25 + 8.00 + 3.00}{10} = 4.70$. The sample standard deviation (s) calculation involves more steps, but once calculated, we use the t-distribution because the population standard deviation is unknown and the sample size is small (n=10). The formula for the confidence interval is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the t-value for 95% confidence and 9 degrees of freedom (n-1). Plugging the values into this formula gives us the confidence interval ($3.39, $6.01), assuming correct calculations for mean, standard deviation, and the use of the appropriate t-value.

A Random Sample of 10 Parking Meters in a Beach $3.60$4.50$2.80$6.30$2.60$5.20$6.75$4.25$8.00$3.00\$ 3.60 \quad \$ 4.50 \quad \$ 2.80 \quad \$ 6.30 \quad \$ 2.60 \quad \$ 5.20 \quad \$ 6.75 \quad \$ 4.25 \quad \$ 8.00 \quad \$ 3.00$3.60$4.50$2.80$6.30$2.60$5.20$6.75$4.25$8.00$3.00

A Random Sample of 10 Parking Meters in a Beach $\$ 3.60 \quad \$ 4.50 \quad \$ 2.80 \quad \$ 6.30 \quad \$ 2.60 \quad \$ 5.20 \quad \$ 6.75 \quad \$ 4.25 \quad \$ 8.00 \quad \$ 3.00$