Homework 10: Problem 8 (1 point) A sample of 123 was...

Homework 10: Problem 8 (1 point) A sample of 123 was selected from a large population. 69% of the sample had a characteristic of interest, and the confidence interval estimate for the proportion of the population with the characteristic was found to be (0.62953266, 0.75046734). What $z_\alpha$ was used to create this interval estimate? Round to two decimal places. What is the confidence level for this estimate?

Transcript

00:01 Hello students, here we have the township board of meridian wants to know how much public support there is for racing property taxes to fix and maintain the roads in the township.

00:13 And they got 90 % confidence interval, which is 0 .58 to 0 .161.

00:19 So also we have n1 is equal to 255 and n2 is equal to 510.

00:24 So we have to find out this sample proportion.

00:27 There we have three statements here.

00:28 So we have to find out which is the correct statement.

00:31 And also we had to find margin of error, standard error, and we have to construct a 95 percentage confidence in travel.

00:37 Okay.

00:38 So let me start with the first one.

00:40 We have to find the sample proportion.

00:42 So here is the equation upper limit plus lower limit divided by 2.

00:45 Here we have upper limit is 0 .61 plus lower limit is 0 .58 divided by 2.

00:54 And that will be equal to 0 .5.

00:59 0 .5 .0 .595.

01:04 So answer to the part 1, samud proportioning 0 .595.

01:08 In the second question, we have to choose the correct statement.

01:12 And here the correct statement is the option c, a 99 percentage confidence interval, a 99 percentage confidence interval calculated using the same survey data will include more possible values for the population proportion.

01:31 Okay.

01:32 In 99 % confidence interval.

01:36 Here this is the correct statement and it is theoretically proved.

01:40 Okay.

01:40 So we can write...

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