Question

In a survey of 170 randomly selected Ohioans, sixteen said that they were left-handed.

a. Calculate the percentage of the people surveyed who were left-handed, rounded to the nearest hundredth of a percent.

b. Estimate the margin of error, assuming a 95% confidence level. Round to the nearest hundredth of a percent.

c. Determine a 95% confidence interval for the relevant population percentage.

d. Write a confidence statement corresponding to your answer in part c. e. An observer could look at your answer in part c and say that the interval is too wide to give you any useful information. Without manufacturing false data or results, name one thing that could be done to construct a valid confidence interval that is narrower.

Answer 1

a. Number of samples n = 170

Number of left handed is 16

Percentage of left handed people is = = 9.41%

b. Margin error for 95% confidence interval is

ME =

Z = 1.96 for 95% confidence interval,

p=sample proportion 0.0941

Margin Error =

= 0.0439 = 4.39%

c. 95% confidence interval of the population percentage is

= ( 9.41% - 4.39% , 9.41% + 4.39% )

= ( 5.02% , 13.8% )

d. The confidence interval says us that we are 95% confident that the population percentage of left handed people will be between 5.02% and 13.8% .

e. Here we want to narrow the confidence interval . This can be done in two ways .

First , take larger sample size .

If you take larger sample size n, then the margin of error will decrease . Then the width of confidence interval will also decrease.

Second , take lower confidence level .

If you take lower confidence level than 95% , then the z score for the confidence level will decrease . Then the margin of error also decrease and you will get a narrower confidence Interval.