Question

A company randomly surveys 15 VIP customers and records their customer satisfaction scores out of a possible 100 points. Based on the data provided, calculate a 90% confidence interval to estimate the true satisfaction score of all VIP customers.

Answer 1

`0.15 - 1.64\sqrt{(0.15(1-0.15))/(100)}=0.0914`

`0.15 + 1.64\sqrt{(0.15(1-0.15))/(100)}=0.2086`

The 90% confidence interval would be given by (0.0914;0.2086)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64`

The confidence interval for the mean is given by the following formula:

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.15 - 1.64\sqrt{(0.15(1-0.15))/(100)}=0.0914`

`0.15 + 1.64\sqrt{(0.15(1-0.15))/(100)}=0.2086`

The 90% confidence interval would be given by (0.0914;0.2086)