A teacher would like to estimate the mean number of steps students take during the school day. To do so, she selects a random sa
mple of 50 students and gives each one a pedometer at the beginning of the school day. They wear the pedometers all day and then return them to her at the end of the school day. From this, she computes the 98% confidence interval for the true mean number of steps students take during the school day to be 8,500 to 10,200. If the teacher had used a 90% confidence interval rather than a 98% confidence interval, what would happen to the width of the interval
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
It would be 20 percent, I believe. First you would put it into a fraction of 12\60, then you would simplify that into 3\15 then divide as shown. You end up with .2 which would be 20 percent.
