Answer :
To estimate the mean temperature and find the 99% confidence interval, we first need to calculate the sample mean and the sample standard deviation from the given temperatures. The temperatures provided are: 31.7, 38.1, 39.6, 31.6, 20.9, and 0.5 degrees Fahrenheit.
Step 1: Calculate the sample mean ([tex]\bar{x}[/tex])
The sample mean is calculated by adding up all the sample values and dividing by the number of samples:
[tex]\bar{x} = \frac{31.7 + 38.1 + 39.6 + 31.6 + 20.9 + 0.5}{6}[/tex]
[tex]\bar{x} = \frac{162.4}{6}[/tex]
[tex]\bar{x} = 27.07[/tex]
Step 2: Calculate the sample standard deviation ([tex]s[/tex])
The sample standard deviation is calculated as follows:
[tex]s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}[/tex]
Where [tex]x_i[/tex] denotes each sample value, and [tex]n[/tex] is the number of samples.
Calculating [tex]\sum (x_i - \bar{x})^2[/tex]:
[tex]= (31.7 - 27.07)^2 + (38.1 - 27.07)^2 + (39.6 - 27.07)^2 + (31.6 - 27.07)^2 + (20.9 - 27.07)^2 + (0.5 - 27.07)^2[/tex]
[tex]\approx 21.38 + 121.56 + 161.57 + 20.52 + 37.97 + 700.35[/tex]
[tex]= 1063.35[/tex]
Substituting into the standard deviation formula:
[tex]s = \sqrt{\frac{1063.35}{5}}[/tex]
[tex]s = \sqrt{212.67}[/tex]
[tex]s \approx 14.58[/tex]
Step 3: Find the 99% confidence interval
For a 99% confidence interval with a small sample size ([tex]n \leq 30[/tex]), we use the t-distribution. With 5 degrees of freedom ([tex]n-1[/tex]), the t-value for 99% confidence is approximately 4.032.
The margin of error (ME) is calculated as:
[tex]ME = t \times \frac{s}{\sqrt{n}}[/tex]
[tex]ME = 4.032 \times \frac{14.58}{\sqrt{6}}[/tex]
[tex]ME = 4.032 \times 5.95[/tex]
[tex]ME \approx 24.00[/tex]
The confidence interval is:
[tex](\bar{x} - ME, \bar{x} + ME)[/tex]
[tex](27.07 - 24.00, 27.07 + 24.00)[/tex]
[tex](3.07, 51.07)[/tex]
Therefore, the 99% confidence interval for the mean temperature is approximately (3.07, 51.07) degrees Fahrenheit.
