Question

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The 90% confidence interval around this sample mean is

Answer 1

The 90% confidence interval around this sample mean is between $127.51 and $203.29.

Explanation:

We are in posession of the sample's standard deviation, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 17 - 1 = 16

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95([tex]t_(95)`). So we have T = 1.7459

The margin of error is:

M = T*s = 1.7459*21.70 = 37.89

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 165.40 - 37.89 = $127.51

The upper end of the interval is the sample mean added to M. So it is 165.40 + 37.89 = $203.29

The 90% confidence interval around this sample mean is between $127.51 and $203.29.

Answer 2

•Lower confidence limit = $155.2109

•Upper confidence limit ,= $174.5891

Explanation:

We are given:

U = 17

x' = $165.40

S.d = $21.70

a = 10℅ (90℅ confidence interval)

For s.d(x'), we have:

S/√u =
`(21)/(√(7))`

= 5.2630

Therefore, S.E (x') = $5.2630

At a= 0.10 (90℅ confidence level)

and degree of freedom = 16 (17-1)

`Thus, t_1_6_, _0_._1_0 = 1.746`

90℅ confidence interval all around sample mean, will be:

(x' ±
`t_(_n_-_1)`; a S.E (x')

= (165.4±(174.6)(5.2630)

(165.5±9.189)

($156.2109, $175.5891)