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 (b) 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 (c) n 26, t 2.55 a 0.01 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 (d) n 26, t 3.95 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 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 You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Concepts and reason

In general, z-test is used when the sample size is large and standard deviation is known.

T-test is more adaptable than Z-test since Z-test will often require certain conditions to be reliable.

If the population standard deviation $\left( \sigma \right)$ is known, then z test is used.

If the population standard deviation $\left( \sigma \right)$ is unknown, then t test is used.

Fundamentals

Test statistic for one sample t-test is,

$t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}$

The formula of degrees of freedom for $t -$ test is,

$df = n - 1$

Here, $n$ is sample size.

Decision rule:

1.If P-value is less than the Level of significance then reject the null hypothesis, otherwise fail to reject the null hypothesis.

2.If t-test statistic value is greater than the t tabulated value then reject the null hypothesis otherwise fail to reject the null hypothesis.

(a)

From the given information, $n = 15,t = 1.66,\alpha = 0.05$ .

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 15 - 1\\\\ = 14\\\end{array}$

The critical value is,

$\begin{array}{c}\\{t_{\alpha /2}} = {\rm{TINV(0}}{\rm{.05,14)}}\left( {{\rm{Use the excel function}}} \right)\\\\ = 2.145\\\end{array}$

Since the test statistic value is less than the critical value, do not reject the null hypothesis.

(b)

From the given information, $n = 15,t = - 1.66,\alpha = 0.05$ .

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 15 - 1\\\\ = 14\\\end{array}$

The critical value is,

$\begin{array}{c}\\{t_{\alpha /2}} = {\rm{T}}{\rm{.INV(0}}{\rm{.05/2,14)}}\left( {{\rm{Use the excel function}}} \right)\\\\ = - 2.145\\\end{array}$

Since the test statistic value does not lie in the rejection region, do not reject the null hypothesis.

(c)

From the given information, $n = 26,t = - 2.55,\alpha = 0.01$ .

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 26 - 1\\\\ = 25\\\end{array}$

The critical value is,

$\begin{array}{c}\\{t_{\alpha /2}} = {\rm{T}}{\rm{.INV(0}}{\rm{.01/2,25)}}\left( {{\rm{Use the excel function}}} \right)\\\\ = - 2.787\\\end{array}$

Since the test statistic value does not lie in the rejection region, do not reject the null hypothesis.

(d)

From the given information, $n = 26,t = - 3.95,\alpha = 0.01$ .

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 26 - 1\\\\ = 25\\\end{array}$

The critical value is,

$\begin{array}{c}\\{t_{\alpha /2}} = {\rm{T}}{\rm{.INV(0}}{\rm{.01/2,25)}}\left( {{\rm{Use the excel function}}} \right)\\\\ = - 2.787\\\end{array}$

Since the test statistic value lies in the rejection region, reject the null hypothesis.

Ans: Part a

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

Part b

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

Part c

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

Part d

Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in.