Question

learning goal
i can estimate a population mean or proportion given data from a
sample survey. i can determine the margin of error using
simulation models for random sampling.
lesson reflection (circle one)
starting... getting there... got it!
lesson 24.1 checkpoint
once you have completed the above problems and checked your solutions, complete the lesson checkpoint
below.
complete the lesson reflection above by circling your current understanding of the learning goal.

1. in a random sample of 495 four - year - olds in a state, 55% can provide the first and last name of at least one

parent or guardian. find a 99% confidence interval for the population proportion ( p =.55 ). use the
expressions
( hat { p } - z _ { c } sqrt { \frac { hat { p } ( 1 - hat { p } ) } { n } } ) and ( hat { p } + z _ { c } sqrt { \frac { hat { p } ( 1 - hat { p } ) } { n } } )
where ( n ) is the sample size, ( hat { p } ) is the sample proportion, and ( z _ { c } = 2.576 ).
with 99% confidence, the proportion of four - year - olds who can provide the first and last name of at least
one parent or guardian is between:
a. ( 49.2% ) and ( 60.8% )
b. ( 39.2% ) and ( 50.8% )
c. ( 45.2% ) and ( 69.8% )
d. ( 46.4% ) and ( 63.8% )

1. omar manages the security team at a large airport and surveys a random sample of 149 travelers. he finds

that the mean amount of time that it takes passengers to clear security is 28 minutes and that the population
standard deviation is 6.5 minutes. find a 90% confidence interval for the population mean. use the
expressions
( overline { x } - z _ { c } \frac { sigma } { sqrt { n } } ) and ( overline { x } + z _ { c } \frac { sigma } { sqrt { n } } )
where ( n ) is the sample size, ( overline { x } ) is the sample mean, ( sigma ) is the sample population, and ( z _ { c } = 1.645 ).
with 90% confidence, the mean amount of time that it takes passengers to clear security at the airport lies
between:
a. 21.5 and 34.5 minutes
b. 17.3 and 38.9 minutes
c. 27.1 and 28.9 minutes
d. 25.1 and 30.9 minutes

Explanation:

Step1: Solve for the confidence interval in question 1

Given \(n = 495\), \(\hat{p}=0.55\), \(z_{c}=2.576\)
First, calculate \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.55\times(1 - 0.55)}{495}}=\sqrt{\frac{0.55\times0.45}{495}}=\sqrt{\frac{0.2475}{495}}=\sqrt{0.0005}=0.0224\)
Then, \(\hat{p}-z_{c}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=0.55-2.576\times0.0224 = 0.55 - 0.0577=0.4923\approx49.2\%\)
\(\hat{p}+z_{c}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=0.55 + 2.576\times0.0224=0.55+0.0577 = 0.6077\approx60.8\%\)

Step2: Solve for the confidence interval in question 2

Given \(n = 149\), \(\bar{x}=28\), \(\sigma = 6.5\), \(z_{c}=1.645\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{6.5}{\sqrt{149}}\approx\frac{6.5}{12.2066}\approx0.5326\)
Then, \(\bar{x}-z_{c}\frac{\sigma}{\sqrt{n}}=28-1.645\times0.5326=28 - 0.876=27.124\approx27.1\)
\(\bar{x}+z_{c}\frac{\sigma}{\sqrt{n}}=28+1.645\times0.5326=28 + 0.876=28.876\approx28.9\)

Answer:

1. A. \(49.2\%\) and \(60.8\%\)
2. C. \(27.1\) and \(28.9\) minutes