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Vlad1618 [11]
3 years ago
14

Think about a population mean that you may be interested in and propose a hypothesis test problem for this parameter. Gather app

ropriate 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.
Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

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

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