Confidence Intervals for Proportions

A study found that in a sample of 44 subjects, 9 of agree that children under 6 shouldn’t be taken to nice restaurants. Calculate a 95% confidence interval for p.

• (0.105, 0.305)

• (0.085, 0.324)

• (0.117, 0.292)

• (0.048, 0.361)

A study found that in a sample of 44 subjects, 9 of agree that children under 6 shouldn’t be taken to nice restaurants.

The 95% confidence interval for p is (0.082, 0.318). Interpret this interval.

• 95% of the time the proportion of people who believe children under 6 shouldn't be taken to nice restaurants will be between 8.2% and 31.8%.

• The population proportion of people who believe children under 6 shouldn't be taken to nice restaurants is between 8.5% and 32.4%

• The probability that the proportion of people who believe children under 6 shouldn't be taken to nice restaurants will be between 8.2% and 31.8% is 95%.

• Based on this sample, I am 95% confident that the proportion of people who believe children under 6 shouldn't be taken to nice restaurants is between 8.5% and 32.4%

Tags

CCSS.HSS.IC.A.1

CCSS.HSS.IC.B.4

A survey of 121 subjects found that 84 have suffered the symptoms of depression in the last year. Calculate a 90% confidence interval for p.

• (0.612, 0.776)

• (0.641, 0.748)

• (0.780, 0.900)

• (0.621, 0.759)

Tags

CCSS.HSS.IC.B.4

A survey of 121 subjects found that 84 have suffered the symptoms of depression in the last year.

A 90% confidence interval for p is (0.621, 0.759). Interpret this interval.

• Based on this sample, I am 90% confident that the true proportion of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%.

• 90% of the time, the population of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%

• 90% of all of the possible intervals calculated using this method will capture the true proportion of people who have suffered the symptoms of depression in the last year.

• The probability that people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9% is 90%.

A survey of 121 subjects found that 84 have suffered the symptoms of depression in the last year.

A 90% confidence interval for p is (0.621, 0.759). In this context, interpret "90% Confidence".

• Based on this sample, I am 90% confident that the true proportion of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%.

• 90% of the time, the population of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%

• 90% of all of the possible intervals calculated using this method will capture the true proportion of people who have suffered the symptoms of depression in the last year.

• The probability that people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9% is 90%.

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)

Tags

CCSS.HSS.IC.A.1

CCSS.HSS.IC.B.4

CCSS.HSS.ID.A.4

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

Tags

CCSS.HSS.IC.A.1

CCSS.HSS.IC.B.4

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 "99% confidence."

• 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

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 "99% confidence."

• 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

A preliminary study suggests that 17% of people agree on an issue. How large must your sample size be so you could be within ± 5% with 95% confidence?

• 217

• 30

• 2

• 152

Tags

CCSS.HSS.IC.A.1

CCSS.HSS.IC.B.4