Question

1. A survey of 200 students is selected randomly on a large university campus They are asked if they use a laptop in class to take notes. The result of the survey is that 70 of the 200 students use laptops to take notes in class.
   1. What is the value of the sample proportion?
      
   2. What is the standard error of the sampling proportion?
      
   3. Construct an approximate 95% confidence interval for the true proportion by going 2 standard errors from the sampling proportion.
      
   4. How would the confidence interval have changed if the confidence level was 90% instead of 95%?

  e. How would the confidence interval have changed if the sample size had been 100 instead of 200?

Answer 1

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a.

Value of sample prop
= pcap
= x/n
= 70/200
= .35

b.

Standard error
= sqrt(pcap*pcap'/n)
= sqrt(.35*.65/200)
= 0.0337

c. 95% CI with z = 2 ( 2 std errors), we have:

Formula: pcap +/- Z*SE
= .35 +/- 2*.0337
= 0.2826 to 0.4174

95% CI is : (0.2826, 0.4174)

d. If you use 90% instead of 95% it would mean lower confidence level, hence the Z will decrease causing the width of confidence interval to decrease

e. If you make sample size smaller from 200 to 100 then standard error will increase ( as SE is sqrt(pcap*pcap'/n) in part b) ). When standard error increases, width of confidence interval increases. ( refer formula above)