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A random sample of size nwill be selected from a population, and the proportion of those in the sample who have a Facebook page will be calculated. How would the margin of error for a 95%confidence interval be affected if the sample size were increased from 50to 200?

(a) It remains the same.

(b) It is multiplied by 2.

(c) It is multiplied by 4.

(d) It is divided by 2.

(e) It is divided by 4.

Short Answer

Expert verified

Step 1: Given information

Step by step solution

01

Given information

We are given that in random sample of size n ,confidence interval of 95%

From given information we need to find that what will be change in margin of error if sample is increased from 50to 200.

02

Explanation

Confidence level is a range of values so defined that there is a specified probability that the value of a parameter lies within it.

Formula of confidence level is

X¯±Zασ/2n

Where

X¯= Mean

= Confidence coefficient

= Confidence level

= Standard deviation

= sample space

The value after the symbol is known as the margin of error.

We need to find change in margin of error(Let it be )

For confidence intervalwe have

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Most popular questions from this chapter

Facebook As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003U.S. adults. In these two studies, 71.0%of teens and 58.0%of adults used Facebook. Let plocalid="1654194806576" T3051526=0.200=20.0%pT= the true proportion of all U.S. teens who use Facebook and pA3051526=0.200=20%pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a 99%confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.

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Based on the P-value in Exercise 71, which of the following must be true?

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b. A 95%confidence interval for μM−μF3051526=0.200=20.0%μM-μFwill contain 0

c. A 99%confidence interval for μM−μF3051526=0.200=20.0%μM-μFwill contain 0

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