Question

- 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.
- What is the value of the sample proportion?
- What is the standard error of the sampling
proportion?
- Construct an approximate 95% confidence interval for the true
proportion by going 2 standard errors from the sampling
proportion.
- How would the confidence interval have changed if the confidence level was 90% instead of 95%?

- What is the value of the sample proportion?

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

Answer #1

**Please don't hesitate to give a "thumbs up" for the
answer in case the answer has helped you**

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)**

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