Question

taking into account your Instructor's unfair attempts to confuse you)? is the maximum possible numbcr or ths Infomation for Q2-5 You ask 100 students at MassBay if they would anonymously share with you the n The median of these numbers is 20 hours, and the mean 18.2 hrs/wk with a 64% in your sample work at least 22 hours per week currently work while going to school. 81 of these agree and share the roquested info. Q 2 Construct a 96% CI for the percentage of statewide community college studcnts working > 22 hr/wk. Q 3 Unsure how your result in Q 2 will be received you adjust Cl to 86% How is the result differ Q 4 what is the statewide student population mean hrs/wk worked for 90% CI? Q 5 What sample size would you need to get the same E for Q 2 as in Q 3 (same value for E as in Q 3, but at 96% CI)? ent from Q 2 answer?

Answer 1

2. 96% confidence interval for population proportion

$\hat{p}\pm z_{c}*\sqrt{\hat{p}(1-\hat{p})/n}$

For 96% confidence , zc = 2.06

therefore 96% confidence interval is

$0.64\pm 2.06*\sqrt{0.64(1-0.64)/81}$

= (0.5301 , 0.7499)

96% confidence interval for percentage of students who work at least 22 hrs/wk is

(53.01% , 74.99%)

3.

86% confidence interval for population proportion

$\hat{p}\pm z_{c}*\sqrt{\hat{p}(1-\hat{p})/n}$

For 86% confidence , zc = 1.48

therefore 96% confidence interval is

$0.64\pm 1.48*\sqrt{0.64(1-0.64)/81}$

= (0.5611 , 0.7189)

86% confidence interval for percentage of students who work at least 22 hrs/wk is

(56.11% , 71.89%)

86% confidence interval is narrower than 96% confidence interval .

4.90% confidence interval for population mean

$\bar{X}\pm t_{c}*s /\sqrt{n}$

For 90% CI with df = 81-1=80

tc = 1.6641

Therefore 90% CI is

$18.2\pm 1.6641*3.5 /\sqrt{81}$

= (17.55 , 18.85)

5. In Q3.

$E=1.48*\sqrt{0.64(1-0.64)/81}$

= 0.0789

Now for 96% CI , zc =2.06

Therefore

$2.06*\sqrt{0.64(1-0.64)/n}= 0.0789$

$\Rightarrow n = 157.06$

therefore sample size should be 157