A survey of 150 students is selected randomly on a large university campus. They

Question

A survey of 150 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 84 of the 150 students responded 'yes," An approximate 95% confidence nterval is (0 481, 0 639) Complete parts a trough d belo. a) How would the confidence interval change if the confidence level had been 90% nstead of 95%? The new confidence interval would be The new confidence interval would be Round to three decimal places as b) How would the confidence interval change if the sample size had been 375 instead of 150? (Assume the same sample proportion,) The new confidence interval would be The new confidence interval would be (Round to three decimal places as needed) c) How would the confidence interval change if the confidence level had been 98% nstead of 95%? The new confidence interval would be (Round to three decimal places as needed) | The new confidence interval would be

Explanation / Answer

Solution:

We are given

n = 150, x = 84,

So, P = x/n = 84/150 = 0.56

Confidence interval formula is given as below:

Confidence interval = P -/+ Z*sqrt(P(1 – P)/n)

Where, Z is critical value at /2 level of significance.

sqrt(P(1 – P)/n) = sqrt(0.56*(1 – 0.56)/150) = 0.0405

Lower limit = P - Z*sqrt(P(1 – P)/n)

Upper limit = P + Z*sqrt(P(1 – P)/n)

Part a

We are given

Confidence level = 90%

So, Z = 1.6449

Lower limit = P - Z*sqrt(P(1 – P)/n)

Lower limit = 0.56 – 1.6449* 0.0405 = 0.493

Upper limit = 0.56 + 1.6449*0.0405 = 0.627

The new confidence interval would be decreases.

The new confidence interval would be (0.493, 0.627).

Part b

We are given n = 375

sqrt(P(1 – P)/n) = sqrt(0.56*(1 – 0.56)/375) = 0.025633

For 95% confidence level, Z = 1.96

Lower limit = P - Z*sqrt(P(1 – P)/n)

Lower limit = 0.56 – 1.96*0.025633 = 0.509759

Upper limit = 0.56 + 1.96*0.025633 =0.610241

The new confidence interval would be decreases.

The new confidence interval = (0.510, 0.610)

Part c

Now, we have given confidence level = 98%

So, Z = 2.3263

Lower limit = P - Z*sqrt(P(1 – P)/n)

Lower limit = 0.56 – 2.3263* 0.0405 = 0.466

Upper limit = 0.56 + 2.3263* 0.0405 = 0.654

The new confidence interval would be increases.

The new confidence interval would be (0.466, .654).

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