Question

Think about a population mean that you may be interested in and propose a hypothesis test problem for this parameter. Gather appropriate data and post your problem, Later, respond to your own post with your own solution. For example, you may believe that the population mean number of times that adults go out for dinner each week is less than 1.5. Your data could be that you spoke with 7 people and found that they went out 2, 0, 1, 5, 0, 2, and 3 times last week. You then would choose to test this hypothesis at the .05 (or another) significance level. Assume a random sample.
Required:
Test this hypothesis at the 0.05 significance level.

Answer 1

Answer:

There is no sufficient evidence to conclude that population mean number of times that adults go out for dinner each week is less than 1.5

Step-by-step explanation:

From the question we are told that

   The  population mean is  less than  1.5

   The sample size is  n  =  7

   The sample data is   2, 0, 1, 5, 0, 2, and 3

Generally the sample mean is mathematically represented as  

     $\= x = \frac{ \sum x_i }{ n }$

=>  $\= x = \frac{ 2 + 0 + 1 + 5 + 0 + 2 + 3 }{ 7 }$

=>  $\= x = 1.857$

Generally the standard deviation is mathematically represented as

      $\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

=>   $\sigma = \sqrt{\frac{ (2 - 1.857)^2 + (0 - 1.857)^2 +\cdots (3 - 1.857)^2}{7} }$

=>   $\sigma = 1.773$

The null hypothesis is  $H_o : \mu = 1.5$

The alternative hypothesis is $H_a : \mu < 1.5$

Generally the test statistics is mathematically represented as

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

=>   $z = \frac{\= x - 1.857 }{ \frac{1.773 }{\sqrt{7} } }$  

=>$z = 0.53$

Generally the p-value is mathematically represented as

      $p-value = P(z < 0.53 )$

From the z-table  

        $P(z < 0.53 ) = 0.70194$

So

       $p-value = 0.70194$

From the values obtained we see that $p-value > 0.05$  

Here there is no sufficient evidence to conclude that population mean number of times that adults go out for dinner each week is less than 1.5