A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation
for the sample was 25 pounds. Find the 99% confidence interval of the population mean for the
weights of adult elephants. Assume the variable is normally distributed. Round intermediate
answers to at least three decimal places. Round your final answers to the nearest whole number.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of . So we have T = 3.25

The margin of error is:

M = T*s = 3.25*25 = 81

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds

The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

really really sorry if i'm

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