Confidence Intervals Proportion

A recent study of 750 Internet users in Europe found that 35% of Internet users were women.  What is the 95% confidence interval of the true proportion of women in Europe who use the Internet? 

• 0.321 < p < 0.379

• 0.316 < p < 0.384

• 0.309 < p < 0.391

• 0.305 < p < 0.395

A survey of 800 women shoppers found that 17% of them shop on impulse.  What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse? 

• 0.148 < p < 0.192

• 0.136 < p < 0.204

• 0.144 < p < 0.196

• 0.139 < p < 0.201

It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets.  What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets? 

• (0.621, 0.780)

• (0.591, 0.812)

• (0.584, 0.830)

• (0.615, 0.805)

Which of the following actions would produce a confidence interval with a smaller width assuming all other contingencies remained constant?

• Using a higher confidence level

• Using a lower confidence level

Sanjay is researching if female students are more or less likely than male students to have received extra credit at a large university. He takes a random sample of 300 students. Sanjay used this sample to build a 95% confidence interval to estimate the difference between the proportion of females and males receiving extra credit. The resulting interval was 0.05±0.07. Based on the interval, what do we know about the corresponding P-value and conclusion at the α=0.05 level of significance?

• The P-value is less than α=0.05, and he cannot conclude that there is a difference between the proportions.

• The P-value is less than α=0.05, and he should conclude that there is a difference between the proportions.

• The P-value is greater than α=0.05, and he cannot conclude that there is a difference between the proportions.

• The P-value is greater than α=0.05, and he should conclude that there is a difference between the proportions.

Sanjay is researching if female students are more or less likely than male students to have received extra credit at a large university. He takes a random sample of 300 students. Sanjay used this sample to build a 95% confidence interval to estimate the difference between the proportion of females and males receiving extra credit. The resulting interval was 0.05±0.07. Based on the interval, what do we know about the corresponding P-value and conclusion at the α=0.05 level of significance?

• The P-value is less than α=0.05, and he cannot conclude that there is a difference between the proportions.

• The P-value is less than α=0.05, and he should conclude that there is a difference between the proportions.

• The P-value is greater than α=0.05, and he cannot conclude that there is a difference between the proportions.

• The P-value is greater than α=0.05, and he should conclude that there is a difference between the proportions.

Which of the following changes to a study would result in a narrower confidence interval?

• increasing the confidence level, increasing the sample size

• decreasing the confidence level, decreasing the sample size

• increasing the confidence level, decreasing the sample size

• decreasing the confidence level, increasing the sample size.

A researcher at a major hospital wishes to estimate the proportion of the adult population of the U.S. that has high blood pressure.  How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%?

• 9

• 271

• 164

• 542

In a survey of 104 students, it was found that 79 went to the homecoming game this year. Calculate a 99% confidence interval for p.

• (0.652, 0.868)

• (0.691, 0.829)

• (0.678, 0.842(

• (0.685, 0.895)

In a survey of 104 students, it was found that 79 went to the homecoming game this year.

A 99% confidence interval for p is (0.652, 0.868). Interpret this interval.

• 99% of the time the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.

• The probability that the population proportion of people who went to the homecoming game this year is between 65.2% and 86.6% is 95%.

• Based on this sample, I am 99% confident that the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.

• 99% of all possible intervals calculated this way will capture the true proportion of people who went to the homecoming game this year