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The high temperatures (in degrees Fahrenheit) of a random sample of 6 small towns are: 98.9, 98.5, 96.5, 98.8, 97.8, 99.8.

Assume high temperatures are normally distributed. Based on this data, find the 99% confidence interval of the mean high temperature of the towns.

Enter your answer as an open interval (i.e., parentheses), accurate to two decimal places. The sample data are reported accurate to one decimal place. The answer should be obtained without any preliminary rounding; however, the critical value may be rounded to 3 decimal places.

Answer :

We can be 99% confident that the true average high temperature of all towns is between 96.53 and 100.24 degrees Fahrenheit.

To solve this question, we need to follow a few mathematical steps to find the 99% confidence interval for the mean high temperature:
1. Firstly, we list out the given temperatures, which are 98.9, 98.5, 96.5, 98.8, 97.8, and 99.8 degrees Fahrenheit. These temperatures come from a random sample of 6 towns.
2. Next, we find the mean of these temperatures. This is done by adding up all temperatures and dividing by the total number of temperatures. After performing this calculation, we find that the mean temperature is approximately 98.38 degrees Fahrenheit.
3. To measure how spread out the data is, we calculate the standard deviation. In our case, the standard deviation is approximately 1.127.
4. To find the 99% confidence interval, we first need to determine the critical value. This is found using the T-distribution, due to our small sample size. The critical value in our case is approximately 4.032 because we have a 6-sample size.
5. Now, we calculate the margin of error, which tells us the range in which we expect our population mean would fall based on our sample. The margin of error is found by multiplying the critical value by the standard deviation divided by the square root of the sample size. Our calculated margin of error is approximately 1.855.
6. Finally, to find the 99% confidence interval, we subtract the margin of error from the mean and add the margin of error to the mean. This gives us an interval, which is (96.53, 100.24).

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