Question: The high temperatures (in degrees Fahrenheit) of a random sample of 7 small towns are:
99.3
96.6
96.3
99.4
97.1
96.5
98.5
Assume high temperatures are normally distributed. Based on this data, find the 98% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).
98% C.I. =
1. **State the problem:** We have a sample of $n=7$ high temperatures: $99.3, 96.6, 96.3, 99.4, 97.1, 96.5, 98.5$. We want to find the 98% confidence interval for the population mean assuming normal distribution.
2. **Formula for confidence interval:** Since the population standard deviation is unknown and the sample size is small, we use the $t$-distribution:
$$\text{CI} = \left(\bar{x} - t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}, \ \bar{x} + t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} \right)$$
where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2, n-1}$ is the critical $t$-value for confidence level $1-\alpha$ with $n-1$ degrees of freedom.
3. **Calculate the sample mean $\bar{x}$:**
$$\bar{x} = \frac{99.3 + 96.6 + 96.3 + 99.4 + 97.1 + 96.5 + 98.5}{7} = \frac{683.7}{7} = 97.6714285714$$
4. **Calculate the sample standard deviation $s$:**
First, find squared deviations:
\begin{align*}
(99.3 - 97.6714)^2 &= 2.6467 \\
(96.6 - 97.6714)^2 &= 1.1467 \\
(96.3 - 97.6714)^2 &= 1.8767 \\
(99.4 - 97.6714)^2 &= 2.9971 \\
(97.1 - 97.6714)^2 &= 0.3271 \\
(96.5 - 97.6714)^2 &= 1.3710 \\
(98.5 - 97.6714)^2 &= 0.6867
\end{align*}
Sum of squared deviations:
$$2.6467 + 1.1467 + 1.8767 + 2.9971 + 0.3271 + 1.3710 + 0.6867 = 11.0519$$
Sample variance:
$$s^2 = \frac{11.0519}{7-1} = \frac{11.0519}{6} = 1.8419833$$
Sample standard deviation:
$$s = \sqrt{1.8419833} = 1.3561$$
5. **Find the critical $t$-value for 98% confidence and $df=6$:**
From $t$-tables or calculator, $t_{0.01,6} = 3.143$ (rounded to 3 decimals).
6. **Calculate margin of error:**
$$E = t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} = 3.143 \times \frac{1.3561}{\sqrt{7}} = 3.143 \times \frac{1.3561}{2.6458} = 3.143 \times 0.5125 = 1.610$$
7. **Construct confidence interval:**
$$\left(97.6714 - 1.610, \ 97.6714 + 1.610 \right) = (96.06, 99.28)$$
8. **Round to two decimal places:**
$$\boxed{(96.06, 99.28)}$$
This is the 98% confidence interval for the mean high temperature of the towns based on the sample data.