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?
**(0.5 pts)** - What is the standard error of the sampling proportion?
**(0.5 pts)** - Construct an approximate 95% confidence interval for the true
proportion by going 2 standard errors from the sampling proportion.
**(1 pt)** - How would the confidence interval have changed if the
confidence level was 90% instead of 95%?
**(1 pt)** - How would the confidence interval have changed if the sample
size had been 100 instead of 200?
**(1 pt)**

- What is the value of the sample proportion?

Answer #1

From the given information,

The required correct answers are,

a. Sample proportion(p)= 70/200 = 0.35

b. Standard error= Squareroot((0.35*(1-0.35))/200)= 0.0337

c. By using calculator,

Confidence Interval= (0.35-1.96*0.0337,0.35+1.96*0.0337)

Confidence Interval= (0.2839,0.4161)

d. We know that, As confidence level decreases, confidence interval also decreases.

Hence, confidence interval will decrease.

e. We know that, As sample size decreases, confidence interval increases.

Hence, confidence interval will increase.

Thank you.

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