Question

lesson 23.3 and 24.1
estimate a population mean or proportion given data from a
how i did (circle one)
sample survey. i can determine the margin of error using simulation
i got it!
im still learning it.
models for random sampling.
about 40% of athletes at a gym use a professional trainer. the proportion of successes is when ( p = 0.4 ). you
choose a random sample of 50 athletes at the gym. use this information to answer questions 14 and 15.

1. what is the standard error of the mean? use ( sigma_{p}=sqrt{\frac{p(1 - p)}{n}} ). (lesson 23.3) (1 point)

a. 0.110
b. 0.089
c. 0.069

1. what interval captures ( 68% ) of the proportions for random samples of 50 athletes? use ( sigma_{p}=sqrt{\frac{p(1 - p)}{n}} )

(lesson 23.3) (1 point)
a. between ( 29% ) and ( 51% )
b. between ( 33.1% ) and ( 46.9% )
c. between ( 31.1% ) and ( 48.9% )
use the table below for problem 16.

1. in a random sample of 80 parents in the united states, ( 60% ) used the social media app instagram. (lesson

24.1)
find a ( 95% ) confidence interval for the population proportion ( p ). use the expressions
hat{p}-z_{c} sqrt{\frac{hat{p}(1-hat{p})}{n}} \text { and } hat{p}+z_{c} sqrt{\frac{hat{p}(1-hat{p})}{n}}
where ( n ) is the sample size, ( p ) is the sample proportion, and ( z_{c}=1.96 ).
with ( 95% ) confidence, the proportion for all parents in the united states who use
instagram lies between: (1 point)
a. ( 43% ) and ( 77% )
b ( 54.5% ) and ( 65.5% )
c. ( 46.1% ) and ( 73.9% )
d. ( 49.3% ) and ( 70.7% )

Explanation:

Step1: Calculate standard error for question 14

Given \(p = 0.4\), \(n=50\). Use the formula \(\sigma_p=\sqrt{\frac{p(1 - p)}{n}}\).
Substitute values: \(\sigma_p=\sqrt{\frac{0.4\times(1 - 0.4)}{50}}=\sqrt{\frac{0.4\times0.6}{50}}=\sqrt{\frac{0.24}{50}}=\sqrt{0.0048}\approx0.069\)

Step2: Calculate interval for question 15

For a normal distribution, about \(68\%\) of the data lies within \(p\pm\sigma_p\).
\(p = 0.4\) (or \(40\%\)), \(\sigma_p\approx0.069\) (or \(6.9\%\)).
Lower bound: \(40\% - 6.9\%=33.1\%\)
Upper bound: \(40\%+6.9\% = 46.9\%\)

Step3: Calculate confidence interval for question 16

Given \(\hat{p}=0.6\), \(n = 80\), \(z_c=1.96\)
First, calculate \(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.6\times(1 - 0.6)}{80}}=\sqrt{\frac{0.6\times0.4}{80}}=\sqrt{\frac{0.24}{80}}=\sqrt{0.003}\approx0.055\)
Lower bound: \(\hat{p}-z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.6-1.96\times0.055 = 0.6 - 0.1078=0.4922\approx49.22\%\)
Upper bound: \(\hat{p}+z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.6 + 1.96\times0.055=0.6+0.1078 = 0.7078\approx70.78\%\)

Answer:

1. C. 0.069
2. B. Between 33.1% and 46.9%
3. D. 49.3% and 70.7%