A 100(1 − α)% confidence interval for the mean μ of a normal population when the value of σ is known is given by x − zα/2 · σ n , x + zα/2 · σ n . By how much must the sample size n be increased if the width of the CI above is to be halved? Halving the length requires n to be increased by a factor of . If the sample size is increased by a factor of 25

And we want to reduce this width by a factor of 1/2 so the new margin of error would be:

And in order to do this we need to increase the sample size by a factor of 4 since:

Halving the length requires n to be increased by a factor of 4

If the sample zise is increased by a factor of 25 we got:

the width is reduced by a factor of 5

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we know the the confidence interval for the mean is btween these two limits:

And for this case the width is given by:

And we want to reduce this width by a factor of 1/2 so the new margin of error would be:

And in order to do this we need to increase the sample size by a factor of 4 since:

Halving the length requires n to be increased by a factor of 4

If the sample zise is increased by a factor of 25 we got:

the width is reduced by a factor of 5

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.