Assuming that the sample proportion does not change, what would be the relationship between the width of the original confidence interval and the width of a second 90 percent confidence interval that is created based on a sample of only 1,000 viewers in the area

Complete question:

A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor moving the broadcast of the late weeknight news to an hour earlier than it is currently. Initially, the confidence interval will be created using a simple random sample of 9,000 viewers in the area. Assuming that the sample proportion does not change, what would be the relationship between the width of the original confidence interval and the width of a second 90 percent confidence interval that is created based on a sample of only 1,000 viewers in the area?

original interval is 3 times narrower than the second interval with a sample. Size of 1000

Step-by-step explanation:

The width of the two different intervals, created with sample sizes of 9000 and 1000 samples :

Width = 2z√(pq)/n

q = 1 -p

For :

The only difference between the intervals Is the sample size, n

For Interval1, with sample size, n = 9000 ;

2z√(pq)/9000 ---(1)

For Interval1, with sample size, n = 1000 ;

2z√(pq)/1000 (2)

Dividing 1 by 2 :

2z√(pq)/9000 ÷ 2z√(pq)/1000

2z√(pq)/9000 * 1000 /2z√(pq)

Hence, we have :

√9000 / √1000

300 / 100

= 3

This means the original interval is 3 times narrower than the second interval with a sample. Size of 1000 ; the width of confidence interval narrows as sample size increases

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

sine

step-by-step explanation: