A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation
1 answer:
Answer:
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.
You might be interested in
All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is
Multiply both sides by <em>r</em> :
Subtract the latter sum from the first, which eliminates all but the first and last terms:
Solve for :
Then as gets arbitrarily large, the term will converge to 0, leaving us with
So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18