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Sav [38]
3 years ago
14

Megan and her little sister, Nora, counted their Halloween candy after trick-or-treating. They love taffy and wanted to know who

collected more taffy pieces compared to the rest of the candy. In Megan's bag, 20 out of the 50 candy pieces were taffy. In Nora's bag, 11 out of the 25 pieces were taffy. Who got the greater ratio of taffy to total candy pieces?
A.
Megan got the greater ratio
B.
Nora got the greater ration.
C.
neither. the girls got the same ration
Mathematics
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

A. Megan got the greater ratio

Step-by-step explanation:

the numbers are bigger than Nora's numbers

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The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t
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Complete Question

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  

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        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

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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

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            df =  n-  1

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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

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        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

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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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