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Nimfa-mama [501]
3 years ago
15

I collect a random sample of size n from a population and compute a 95% confidence interval for the proportion I observe from th

e population. What could I do to produce a new confidence interval with a smaller width, smaller margin of error, based on these same data?
Mathematics
1 answer:
adoni [48]3 years ago
3 0

Answer:

You should increase the size of your random sample, that is, increase the value of n.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

This means that as we increase the sample size(n), the margin of error decreases, as does the width of the confidence interval.

What could I do to produce a new confidence interval with a smaller width, smaller margin of error, based on these same data?

You should increase the size of your random sample, that is, increase the value of n.

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Jason and Jill or two students in mr. White's math class. On the last 5 quizzes Jason scored an 80 90 95 85 and 70. Jill scored
Mrac [35]

Jason's scores : 80 90 95 85 70 and

Jill's score : 70 75 90 100 95.

Mean of Jason's scores = \frac{80 + 90 + 95 + 85 + 70}{5}=\frac{420}{5}=84.

Mean of Jill's scores = \frac{70+75+90+100+95}{5}=\frac{430}{5}=86

Now, in order to find the mean absolute deviation, need to find the difference of each score from means.

<u>Mean absolute deviation for Jason's scores.</u>        

|84-80| = 4

|84-90| = 6

|84-95| = 9

|84-85| = 1

|84-70|= 14

\frac{4+6+9+1+14}{5}=\frac{34}{5}=6.8

<u>Mean absolute deviation for Jill's scores</u>

|86-70| = 16

|86-75| = 11

|86-90| = 4

|86-100| = 14

|86-95|= 9

\frac{16+11+4+14+9}{5}=\frac{54}{5}=10.8

Jill got average quiz score 86 and Jason got 84.

Therefore, Jill got better quiz average.

Also, the mean absolute deviation for Jason scores is less that is 6.8 than 10.8.

Therefore, Jason got more consistent grades.

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