The true average diameter of ball bearings of a certain type is supposed to be 0.05 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=13, t=1.6, α=0.05.
b. n=13, t=−1.6, α=0.05.
c. n=25, t=−2.6, α=0.01.

Respuesta :

Answer:

a) H0: [tex] \mu \leq \mu_o[/tex]

H1: [tex] \mu > \mu_o[/tex]

n = 13 represent the sample size

[tex] t = 1.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 13-1=12[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{12} >1.6) = 0.068[/tex]

Since [tex] p_v >\alpha[/tex] we fail to reject the null hypothesis on this case at 5% of significance.

b) H0: [tex] \mu \geq \mu_o[/tex]

H1: [tex] \mu < \mu_o[/tex]

n = 13 represent the sample size

[tex] t = -1.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 13-1=12[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{12} <-1.6) = 0.068[/tex]

Since [tex] p_v >\alpha[/tex] we fail to reject the null hypothesis on this case at 5% of significance.

c) H0: [tex] \mu \geq \mu_o[/tex]

H1: [tex] \mu < \mu_o[/tex]

n = 25 represent the sample size

[tex] t = -2.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 25-1=24[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{24} <-2.6) = 0.0078[/tex]

Since [tex] p_v <\alpha[/tex] we can reject the null hypothesis on this case at 1% of significance.

Step-by-step explanation:

Part a

For this case we assume that we are testing the following system of hypothesis

H0: [tex] \mu \leq \mu_o[/tex]

H1: [tex] \mu > \mu_o[/tex]

n = 13 represent the sample size

[tex] t = 1.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 13-1=12[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{12} >1.6) = 0.068[/tex]

Since [tex] p_v >\alpha[/tex] we fail to reject the null hypothesis on this case at 5% of significance.

Part b

For this case we assume that we are testing the following system of hypothesis

H0: [tex] \mu \geq \mu_o[/tex]

H1: [tex] \mu < \mu_o[/tex]

n = 13 represent the sample size

[tex] t = -1.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 13-1=12[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{12} <-1.6) = 0.068[/tex]

Since [tex] p_v >\alpha[/tex] we fail to reject the null hypothesis on this case at 5% of significance.

Part c

For this case we assume that we are testing the following system of hypothesis

H0: [tex] \mu \geq \mu_o[/tex]

H1: [tex] \mu < \mu_o[/tex]

n = 25 represent the sample size

[tex] t = -2.6[/tex] represent the calculated statistic

The degrees of freedom are given by:

[tex] df = n-1 = 25-1=24[/tex]

We can calculathe the p value with this formula:

[tex] p_v = P(t_{24} <-2.6) = 0.0078[/tex]

Since [tex] p_v <\alpha[/tex] we can reject the null hypothesis on this case at 1% of significance.