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

Answer:

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

Step-by-step 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  $H_a : \mu \ne 108.50$

Generally the test statistics is mathematically represented as

       $z = \frac{\= x - \mu }{ \frac{ \sigma}{\sqrt{n} } }$

=>    $z = \frac{112 - 108.50 }{ \frac{ 16}{\sqrt{64} } }$  

=>    $z = 1.75$

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