4n^2 -15n+9 factor completely
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Answer:
The margin of error for the 95% confidence interval is of 1.389 ounces per hour.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 100 - 1 = 99
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.9842
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
Then
The margin of error for the 95% confidence interval is of 1.389 ounces per hour.