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 o Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in o Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in o Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in o Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in

Answer 1

The null hypothesis must be rejected as there is enough evidence that the diameter differs from 0.5 inches.

Given values are

The population mean is  0.5 inches

The Null hypothesis is  : H₀ : μ = 0.5 inches

The Alternative hypothesis is  given by Hₐ : μ ≠ 0.5 inches

Now let us consider the parameter given:

sample size is  n =  15  

test statistics is  t =  1.66

The level of significance : α = 0.05

The degree of freedom: df = n-1 = 15 -1 =14

Calculating the critical value : $t_\frac{\alpha}{2} ,df$ = 2.145

We are using the critical value of $t_\frac{\alpha}{2}$ as the test is one-tailed .

Comparing the value of  t and $t_\frac{\alpha}{2}$ we can see that : $t < t_\frac{\alpha}{2}$

Hence the null hypothesis can be rejected.

Given that the value of t does not fall within the range covered by (i.e., the range from -2.145 to the bottom of the normal distribution curve), we are unable to reject the null hypothesis.

To learn more about null hypothesis visit:

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