Question

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 to see whether this is the case. What conclusion is appropriate in each of the following situations?
(a) n 15 t 1.66 a 0.05
A. Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
B. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
C. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
D. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
(b) n 15 t 1.66 a 0.05
Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
(c) n 26, t 2.55 a 0.01
A. Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
B. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
C. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
D. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
(d) n 26, t 3.95
A. Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in
B. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
C. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
D. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   C

b

    C

c

    C

d  

     A

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