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jeyben [28]
3 years ago
9

Stephen needs to place several cubic packages in a shipping box that measures 12 inches by 18 inches by 24 inches. Each cubic pa

ckage has a dimension of 2 inches on each side. Which is closest to the maximum number of cubes that can fit into the box?
200

800

400

650
Mathematics
1 answer:
jeka57 [31]3 years ago
5 0
12÷2=6
18÷2=9
24÷2=12
6×9×12=648
648=650(closest answer)
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The complete question is shown on the first uploaded image

Answer:

a

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b

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c

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d  

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Step-by-step explanation:

From the question we are told that

    The population mean is  \mu  =  0.5 \  in

     

Generally the Null hypothesis is  H_o  :  \mu = 0. 5 \ in

                The Alternative hypothesis is  H_a  :  \mu  \ne  0.5 \ in

Considering the parameter given for part a  

       The sample size is  n =  15  

        The  test statistics is  t =  1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  2.145

We are making use of this  t_{\frac{\alpha }{2} } because it is a one-tail test

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that  t < t_{\frac{\alpha }{2}  } so the null hypothesis would not be rejected

Considering the parameter given for part b  

       The sample size is  n =  15  

        The  test statistics is  t =  -1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  -2.145

Looking at the value of  t and t_{\frac{\alpha}{2} ,df } the we see that t does not lie in the area covered by  t_{\frac{\alpha}{2}  , df } (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

 

Considering the parameter given for part  c

       The sample size is  n =  26  

        The  test statistics is  t =  -2.55

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  2.787

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that t does not lie in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part  d

       The sample size is  n =  26  

        The  test statistics is  t =  -3.95

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  -2.787

Looking at the value of  t and t_{\frac{\alpha}{2}  } the we see that  t  lies in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we  reject the null hypothesis

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3 years ago
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