Question

Which of the following t-scores is correct given the following information? A popular Nordic furniture company is testing a new format for instructions for how to assemble their products. Two groups of individuals are randomly selected with the first group receiving the old format instructions and the second group receiving the new format instructions. Times to completion of an entertainment center for Group 1 (n = 4) were: 2.0, 4.0, 3.5, and 2.5 hours. Times to completion of an entertainment center for Group 2 (n = 4) were: 1.5, 3.0, 2.5, and 1.0 hours.

Answer 1

We use two independent t test to test whether there is significant difference between mean completion time of group 1 with old format instructions and mean completion time of group 2 with new format instructions.

Group 1 $(x_{1})$	Group 2 $(x_{2})$
2.0	1.5
4.0	3.0
3.5	2.5
2.5	1.0

CALCULATION FOR SAMPLE MEAN:

Mean of sample or group 1 $(x_{1}_{bar})=(2.0+4.0+3.5+2.5)/4$

  $=3$

Mean of sample or group 2 $(x_{2}_{bar})=(1.5+3.0+2.5+1.0)/4$

$=2$

CALCULATION FOR SAMPLE STANDARD DEVIATION:

Sample size of group 1 $(n_{1})=4$

Standard deviation of group 1 $(s_{1})=[\sum (x_{1}-x_{1}_{bar})^{2}/(n_{1}-1)]^{1/2}$

$x_{1}$	$(x_{1}-x_{1}_{bar})$	$(x_{1}-x_{1}_{bar})^{2}$
2.0	-1	1
4.0	1	1
3.5	0.5	0.25
2.5	-0.5	0.25

Thus

$(s_{1})=[\sum (x_{1}-x_{1}_{bar})^{2}/(n_{1}-1)]^{1/2}$

$=[2.5/3]^{1/2}$

$=0.9129$

Sample size of group 2 $(n_{2})=4$

Standard deviation of group 2 $(s_{2})=[\sum (x_{2}-x_{2}_{bar})^{2}/(n_{2}-1)]^{1/2}$

$x_{2}$	$(x_{2}-x_{2}_{bar})$	$(x_{2}-x_{2}_{bar})^{2}$
1.5	-0.5	0.25
3.0	1	1
2.5	0.5	0.25
1.0	-1	1

Thus

$(s_{2})=[\sum (x_{2}-x_{2}_{bar})^{2}/(n_{2}-1)]^{1/2}$

$=[2.5/3]^{1/2}$

$=0.9129$

HYPOTHESIS:

$H_{0}:\mu _{1}-\mu _{2}=0$ (That is, there is no significant difference between mean completion time of group 1 receiving old format instructions and mean completion time of group 2 receiving new format instructions.)

$H_{1}:\mu _{1}-\mu _{2}\neq 0$ (That is, there is significant difference between mean completion time of group 1 receiving old format instructions and mean completion time of group 2 receiving new format instructions.)

LEVEL OF SIGNIFICANCE: $\alpha =0.05$

TEST STATISTIC:

$t=[(x_{1}_{bar}-x_{2}_{bar})-(\mu _{1}-\mu _{2})]/[(((n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2})/(n_{1}+n_{2}-2))*((1/n_{1})+(1/n_{2}))]^{1/2}$

which follows t distribution with $(n_{1}+n_{2}-2)$ degrees of freedom.

CALCULATION:

$t=[(x_{1}_{bar}-x_{2}_{bar})-(\mu _{1}-\mu _{2})]/[(((n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2})/(n_{1}+n_{2}-2))*((1/n_{1})+(1/n_{2}))]^{1/2}$

  $=[3-2]/[(((4-1)0.9129^{2}+(4-1)0.9129^{2})/(4+4-2))*((1/4)+(1/4))]^{1/2}$

  $=1/[0.8334*0.5]^{1/2}$

  $=1/0.6455$

$t=1.55$

The calculated t score is $1.55$.

CRITICAL VALUE:

The two tailed t critical value at significance level $\alpha =0.05$ with $(n_{1}+n_{2}-2)=6$ degrees of freedom is $2.447$.

DECISION RULE:

$Reject \, \, H_{0}\, \, if\, \, t\geq 2.447\, \, or\, \, if\, \, t\leq -2.447$

INFERENCE:

Since the calculated t statistic value (1.55) is less than the t critical value (2.447), we fail to reject null hypothesis and conclude that there is no significant difference between mean completion time of group 1 receiving old format instructions and mean completion time of group 2 receiving new format instructions.