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Estimating a Population Proportion Answers

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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
164692

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
164702

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
164703

✔ 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
164704

(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)
164705

(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)
164706

✔ who have a computerwho had a computerwho did have a computerin the sample who had a computer

A
who have a computer
B
who had a computer
C
who did have a computer
D
in the sample who had a computer
164707

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
164708

✔ 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
164709

(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)
164710

(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)
164711

✔ that containthat containedthat did containin the sample that contained

A
that contain
B
that contained
C
that did contain
D
in the sample that contained

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