Question

In order to determine the average price of hotel rooms in Atlanta, a sample of 64 hotels was selected. It was determined that the average price of the rooms in the sample was $112. The population standard deviation is known to be $16. Use a .05 level of significance and determine whether or not the average room price is significantly different from $108.50.

Answer 1

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the average room price is significantly different from $108.50

Explanation:

From the question we are told that

The sample size is n = 64

The average price is
`\= x = \$ 112`

The population standard deviation is
`\sigma = \$ 16`

The level of significance is
`\alpha = 0.05`

The population mean is
`\mu = \$ 108.5`

The null hypothesis is
`H_o : \mu = 108.50`

The alternative hypothesis is
`H_a : \mu \\e 108.50`

Generally the test statistics is mathematically represented as

`z = (\= x - \mu )/( ( \sigma)/(√(n) ) )`

=>
`z = (112 - 108.50 )/( ( 16)/(√(64) ) )`

=>
`z = 1.75`

From the z table the area under the normal curve to the left corresponding to 1.75 is

`P( Z > 1.75) = 0.040059`

Generally p-value is mathematically represented as

`p-value = 2 * P( Z > 1.75)`

=>
`p-value = 2 * 0.040059`

=>
`p-value = 0.08012`

From the values obtained we see that
`p-value > \alpha` hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the average room price is significantly different from $108.50