*Also find the Margin of Error for each question* Question 1) Question 4) A depa
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Question
*Also find the Margin of Error for each question*
Question 1)
Question 4)
A department of education reported that in 2007, 68% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 320 individuals who graduated from high school 12 months prior was selected. From this sample, 236 students were found to be enrolled in college or a trade school. Complete parts a through c below a. Construct a 95% confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2013. A 95% confidence interval to estimate the actual proportion has a lower limit of and an upper limit of (Round to three decimal places as needed.)Explanation / Answer
Here in all the questions sample size is large enough to use normal population
I will mention the formulas once and do the calculation for each one of them
Here for each cases we will first find SE(p)=sqrt(p(1-p)/n)
Based on this we will calculate Margin of Error=E=z*SE, Where z is the value corresponding to CI% and can be obtained using z table for corresponding CI%
And finally we will find CI=p+/-E
1. Here p=236/320=0.74
Hence SE(p)=sqrt(0.74*0.26/320)=0.025
So E=z*0.025=1.96*0.025 as for 95% CI P(-1.96<z<1.96)=0.95
So E=0.05
Hence CI=0.74+/-0.048=(0.69,0.79)
2.a. here p=195/275=0.71 and z value is 2.58 asP(-2.58<z<2.58)=0.99
SE(p)=sqrt(0.71*0.29/275)=0.027
E=2.58*0.027=0.07
CI=0.71+/-0.07=(0.64,0.78)
3. Here p=28/400=0.07 and z=2.33 as P(-2.33<z<2.33)=0.98
SE(p)=sqrt(0.07*0.93/400)=0.013
E=2.33*0.013=0.03
Hence CI=0.07+/-0.03=(0.04,0.10)
4. Here p=136/750=0.18 and z value is 1.96 as P9-1.96<z<1.96)=0.95
SE(p)=sqrt(0.18*0.82/750)=0.014
E=1.96*0.014=0.027
CI=0.18+/-0.027=(0.15,0.21)
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