Question

you must estimate the mean temperature (in degrees fahrenheit) with the following sample temperatures:
48
48.6
46.8
5.8
56.4
40.1
26.2
35.1
43.1
find the 99% confidence interval. 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).
99% c.i.=
answer should be obtained without any preliminary rounding.
question help: video post to forum

Explanation:

Step1: Calculate the sample mean ($\bar{x}$)

$$\bar{x}=\frac{48 + 48.6+46.8 + 5.8+56.4+40.1+26.2+35.1+43.1}{9}$$
$$=\frac{359.1}{9}=39.9$$

Step2: Calculate the sample standard deviation ($s$)

First, calculate the deviations from the mean:
\((48 - 39.9)=8.1\), \((48.6 - 39.9)=8.7\), \((46.8 - 39.9)=6.9\), \((5.8 - 39.9)=- 34.1\), \((56.4 - 39.9)=16.5\), \((40.1 - 39.9)=0.2\), \((26.2 - 39.9)=-13.7\), \((35.1 - 39.9)=-4.8\), \((43.1 - 39.9)=3.2\)

Then, calculate the sum of squared deviations:
\(8.1^{2}+8.7^{2}+6.9^{2}+(-34.1)^{2}+16.5^{2}+0.2^{2}+(-13.7)^{2}+(-4.8)^{2}+3.2^{2}\)
\(=65.61+75.69+47.61 + 1162.81+272.25+0.04+187.69+23.04+10.24\)
\(=1844.98\)

The sample variance \(s^{2}=\frac{1844.98}{9 - 1}=\frac{1844.98}{8}=230.6225\)

The sample standard deviation \(s=\sqrt{230.6225}\approx15.19\)

Step3: Determine the critical value ($t_{\alpha/2}$)

The degrees of freedom \(df=n - 1=9-1 = 8\)

For a \(99\%\) confidence interval, \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\)

Using the \(t\)-distribution table or a calculator, \(t_{0.005,8}=3.355\)

Step4: Calculate the margin of error ($E$)

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.355\times\frac{15.19}{\sqrt{9}}\)
\(=3.355\times\frac{15.19}{3}\)
\(=3.355\times5.063\approx16.99\)

Step5: Calculate the confidence interval

The lower limit \(=\bar{x}-E=39.9-16.99 = 22.91\)

The upper limit \(=\bar{x}+E=39.9 + 16.99=56.89\)

Answer:

\((22.91,56.89)\)