Can some help me out
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Answer:
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
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 = 20 - 1 = 19
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.7291
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have . So
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.