Question

Question 6 options:

An online retailer wants to estimate the number of visitors that click on their advertisement from a particular website. Of 978 page views in a day, 8% of the users clicked on the advertisement.


Create a 90% confidence interval for the population proportion of visitors that click on the advertisement.

Answer 1

Answer:
`=(0.06573,\ 0.09427)`

Explanation:

The confidence interval for population proportion (p) is given by :_

`\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

, where n= sample size

z* = Critical value.

`\hat{p}` = Sample proportion.

Let p be the true population proportion of visitors that click on the advertisement.

As per given , we have

n= 978

`\hat{p}=0.08`

Critical value for 90% confidence interval : z* = 1.645 [ By z-table]

Now , 90% confidence interval for the population proportion of visitors that click on the advertisement:

`0.08\pm (1.645) \sqrt{(0.08(1-0.08))/(978)}`

`0.08\pm (1.645) √(0.0000752556237219)`

`0.08\pm (1.645) (0.00867499992633)`

`\approx0.08\pm(0.01427)`

`=(0.08-0.01427,\ 0.08+0.01427)=(0.06573,\ 0.09427)`

Hence, a 90% confidence interval for the population proportion of visitors that click on the advertisement.
`=(0.06573,\ 0.09427)`