AnswersCA-Statistics and Probability BPreparing to Estimate a Population Proportion

Estimating a Population Proportion Answers

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According to a recent random survey of 3,728 US adults, 1,939 report using their cell phones to play online games. Construct and interpret a 95% confidence interval for the proportion of US adults who use their cell phones to play online games. State: confidence interval for the true proportion of US adults who use their cell phones to play online games. Plan: ; Random: Random survey; 10%: 3,728 is less than 10% of all US adults; Large Counts: 1,939 successes and 1,789 failures ≥ 10. Do: Conclude: We are 95% confident that the interval from includes the true proportion of all US adults to play online games.

Answers:
State: confidence interval for the true proportion of US adults who use their cell phones to play online games.:95%
Plan: ; Random: Random survey; 10%: 3,728 is less than 10% of all US adults; Large Counts: 1,939 successes and 1,789 failures ≥ 10.:one-sample z-interval for p
Do::(0.504, 0.536)
Conclude: We are 95% confident that the interval from includes the true proportion of all US adults to play online games.:(0.504, 0.536)
Conclude: We are 95% confident that the interval from includes the true proportion of all US adults to play online games.:who use their cell phones
6

Christian wants to estimate the proportion of seniors who plan to attend her school’s prom. She interviews an SRS of 65 of the 695 seniors in her school and finds that 42 plan to attend the prom. Construct and interpret a 90% confidence interval for the proportion of seniors who plan to attend the prom.

Answers:
State: confidence interval for the true proportion of seniors who plan to attend the prom.:90%
Plan: ; Random: SRS; 10%: 65 is less than 10% of all seniors; Large Counts: 42 successes and 23 failures ≥ 10.:one-sample z-interval for p
Do::(0.549, 0.744)
Conclude: We are 90% confident that the interval from includes the true proportion of all seniors .:(0.549, 0.744)
Conclude: We are 90% confident that the interval from includes the true proportion of all seniors .:who plan to attend prom
7

A nationwide random survey of 1,500 teens aged 13–17 found that approximately 65% have their own desktop or laptop computer. Construct and interpret a 99% confidence interval for the true proportion of teens who have their own desktop or laptop computer.

Answers:
State: confidence interval for the true proportion of teens who have their own desktop or laptop computer.:99%
Plan: ; Random: SRS; 10%: 1,500 is less than 10% of all teens; Large Counts: 975 successes and 525 failures ≥ 10.:one-sample z-interval for p
Do::(0.618, 0.682)
Conclude: We are 99% confident that the interval from includes the true proportion of all teens .:(0.618, 0.682)
Conclude: We are 99% confident that the interval from includes the true proportion of all teens .:who have a computer
8

An inspector at a popcorn factory selects a random sample of 75 bags of popcorn from the hundreds produced each hour and finds that 12 contain at least 10% unpopped kernels. Construct and interpret a 95% confidence interval for the true proportion of popcorn bags that contain at least 10% unpopped kernels.

Answers:
State: confidence interval for the true proportion of popcorn bags that contain at least 10% unpopped kernels.:95%
Plan: ; Random: SRS; 10%: is less than 10% of all bags of popcorn produced; Large Counts: 12 successes and 63 failures ≥ 10.:one-sample z-interval for p
Do::(0.077, 0.243)
Conclude: We are 99% confident that the interval from includes the true proportion of all bags of popcorn at least 10% unpopped kernels.:(0.077, 0.243)
Conclude: We are 99% confident that the interval from includes the true proportion of all bags of popcorn at least 10% unpopped kernels.:that contain
9

Is it reasonable to conclude that more than half of US adults use their cell phones to play online games?

A
Yes, because anyone who has a cell phone is going to use it to play online games.
B
No, because a different sample might give different results.
C
No, because the interval of US adults who use their cell phones to play online games is entirely above 0.5.
D
Yes, because the interval of US adults who use their cell phones to play online games is entirely above 0.5.
10

Is it reasonable to conclude that more than two-thirds of the seniors at Christian’s school plan to attend the prom?

A
No, because a different sample might give different results.
B
No, because the interval of seniors who plan to attend prom is not is entirely above 0.67.
C
Yes, because the interval of seniors who plan to attend prom includes 0.67.
D
Yes, because all the seniors will go to prom. Those who answered no may not have found a date yet.
11

Based on the interval, is it reasonable to conclude that a majority of teens aged 13–17 have their own desktop or laptop computer?

A
No, because the interval of teens who have their own desktop or laptop computer is entirely above 0.5.
B
Yes, because the interval of teens who have their own desktop or laptop computer is entirely above 0.5.
C
No, because a different sample might give different results.
D
Yes, because the sample proportion was 65%.
12

Based on the interval, is it reasonable to conclude that the machine needs to be recalibrated?

A
No, because the interval of popcorn bags that contain at least 10% unpopped kernels is entirely above 0.06.
B
Yes, because the sample proportion was greater than 0.06.
C
Yes, because the interval of popcorn bags that contain at least 10% unpopped kernels is entirely above 0.06.
D
No, because a different sample might give different results.
15

✔ one-sample z-interval for ptwo-sample z-interval for pone-sample t-interval for mutwo-sample t-interval for mu

A
one-sample z-interval for p
B
two-sample z-interval for p
C
one-sample t-interval for mu
D
two-sample t-interval for mu
16

✔ one-sample z-interval for ptwo-sample z-interval for pone-sample t-interval for mutwo-sample t-interval for mu

A
one-sample z-interval for p
B
two-sample z-interval for p
C
one-sample t-interval for mu
D
two-sample t-interval for mu
18

(0.318, 0.382)(0.33, 0.37)✔ (0.618, 0.682)(0.649, 0.651)

A
(0.318, 0.382)
B
(0.33, 0.37)
C
(0.618, 0.682)
D
(0.649, 0.651)
19

(0.09, 0.23)✔ (0.077, 0.243)(0.051, 0.27)(0.037, 0.177)

A
(0.09, 0.23)
B
(0.077, 0.243)
C
(0.051, 0.27)
D
(0.037, 0.177)

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