Find the requested confidence interval.
-A survey revealed the amount of pocket money held by each student in a sample of students from Statistics 205. The amounts (in dollars) are $5.50,10.20,3.15,6.08,4.85,4.80,4.70,8.52,7.90,6.10,3.50$, $5.40,7.10,9.60,6.20$, and $4.28$. Assuming $\sigma = 2.00$, find the $99.74 \%$ confidence interval for the population mean.
A)$\$ 5.62$ to $\$ 6.62$
B) $\$ 4.62$ to $\$ 7.62$
C) $\$ 4.12$ to $\$ 8.12$
D) $\$ 0$ to $\$ 4.00$

Question

Find the requested confidence interval.
-A survey revealed the amount of pocket money held by each student in a sample of students from Statistics 205. The amounts (in dollars) are $5.50,10.20,3.15,6.08,4.85,4.80,4.70,8.52,7.90,6.10,3.50$, $5.40,7.10,9.60,6.20$, and $4.28$. Assuming $\sigma = 2.00$, find the $99.74 \%$ confidence interval for the population mean.  
A)$\$ 5.62$ to $\$ 6.62$
B) $\$ 4.62$ to $\$ 7.62$
C) $\$ 4.12$ to $\$ 8.12$
D) $\$ 0$ to $\$ 4.00$

Quizplus · Accepted Answer

Given the sample size $n = 16$ (since there are 16 data points), the population standard deviation $\sigma = 2.00$, and the confidence level $99.74\%$, which corresponds to roughly 3 standard deviations from the mean in a normal distribution. The sample mean $\bar{x}$ can be calculated as the sum of all sample values divided by the number of samples. $$\bar{x} = \frac{5.50 + 10.20 + 3.15 + 6.08 + 4.85 + 4.80 + 4.70 + 8.52 + 7.90 + 6.10 + 3.50 + 5.40 + 7.10 + 9.60 + 6.20 + 4.28}{16} = \frac{98.38}{16} = 6.14875$$The margin of error (E) for a given confidence level can be calculated using the formula $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the confidence level. For a $99.74\%$ confidence interval, the Z-score is approximately 3 (since $99.74\%$ corresponds to $\pm 3$ standard deviations in a normal distribution).$$E = 3 \cdot \frac{2.00}{\sqrt{16}} = 3 \cdot \frac{2.00}{4} = 3 \cdot 0.5 = 1.5$$Thus, the confidence interval is calculated as:$$\bar{x} \pm E = 6.14875 \pm 1.5 = (4.64875, 7.64875)$$Rounding to two decimal places, the confidence interval is approximately $\$ 4.65$ to $\$ 7.65$, which is not exactly listed among the options. However, the option that closely matches this interval, considering rounding and approximation, is:B) $\$ 4.62$ to $\$ 7.62$