For each of the following test, outline a research scenario and use any hypothetical data to show how the test is conducted. i. One sample t-test ii.

i) One Single t test:
Single mean test is used to compare the mean of a variable in the sample of data to a hypothesized mean in the population from which the data is drawn.
The assumptions for single mean test (one sample t-test) are as follows:
1. The dependent variable must be continuous.
2. The dependent variable should be approximately normally distributed.
3. The observations are independent of one another.
4. The dependent variable should not contain any outliers.
Example:
In 2008, the national average for state and local taxes for a family of four was 4172. A random sample of 40 families in a northeastern state was 4560 with a standard deviation of 1590. At $\alpha =0.05$ level of significance. Is there sufficient evidence to conclude that families in that state pay more state and local taxes than the national average.
Answer:
Null hypothesis:
$\mu =4172$
Alternative hypothesis:
$\mu >4172$
This is a two-tailed test.
Here, the sample mean, barx is 4560.
Population mean, $\mu is4172.$
Sample standard deviation, $s=1590.$
Sample size, n is 40.
The test statistic value can be obtained as follows:
$t=\frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}}$
$t=\frac{4560-4172}{\frac{1590}{\sqrt{40}}}$
Computation of P-value:
The P-value oft-distribution at 39 degrees of freedom can be obtained using the excel formula
$“=1-T.DIST(1.5434,39)”.$
The p-value is 0.0654.
Decision rule:
If p-value $\le \alpha ,$ then reject the null hypothesis. Otherwise, do not reject the null hypothesis.
Conclusion:
Here, p-value 0.0654 is greater than the level of significance (0.05).
Therefore, do not reject the null hypothesis.
There is no sufficient evidence to conclude that the families in that state pay more state and local taxes than the national average.
ii) Independent two samples t test:
The independent samples t-test can be used to examine a possible difference between the means of two independent samples under the following conditions:
1) The selected sample should be a simple random sample from two populations.
2) The samples are independent of one another.
3) The population is approximately normal.
Example:
In the experiment of A and B consumers, 12 sample points are set for A consumers, and the average value of word-of-mouth is 3 with a variance of 1.2, 8 sample points are set for B consumers, and the average value of word-of-mouth is 5 with a variance of 1.6. Compare whether there is significant difference between the average word-of-mouth of A and B consumers. $(\alpha =0.05)$
The null and alternative hypotheses are stated as follows:
$H_0:{\mu }_1-{\mu }_2=0$
$H_1:{\mu }_1-{\mu }_2\le 0$
The pooled variance is computed as follows:
$S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$
$=\frac{11\times 1.2+7\times 1.6}{12+8-2}$
$=1.3556$
The test statistic is computed as follows:
$t=\frac{({\bar{x}}_1-{\bar{x}}_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$
$=\frac{3-5}{\sqrt{1.3556\times (\frac{1}{12}+\frac{1}{8})}}$
$=-3.7634$
Thus, the test statistic value is -3.7634.
Decision rule:
If the p-value is less than or equal to level of significance then reject the null hypothesis.
Otherwise fail to reject the null hypothesis.
Computation of P-value:
The two tailed P-value for the t-testis approximately 0.0014 which can be obtained using the excel formula,$“=2\times (T.DIST(-3.7634,18,1))”.$
Thus, is p-value is 0.0014.
Conclusion:
Assume the level of significance is 0.05.
The P-value is 0.0014 and the significance level is 0.05.
Here, the P-value is less than the significance level.
Hence, the null hypothesis is rejected.
Thus, there is sufficient evidence to conclude that there is significant difference between average word-of-mouth of A and B consumers. iii) Correlated samples t test or paired t test or repeated measures t test:
The paired t is used when the two sample are paired or related samples. When the sample is measured twice, or paired samples are taken, then the two samples would be dependent. This indicates that paired t test is used for comparing the dependent samples.
Conditions to perform the paired t-test:
1) Dependent variable is measured on a continuous scale.
2) Each score in one sample is paired with a particular score in the other sample.
3) The difference of the either group follows normal distribution.
Example:
Junior school research team would like to examine whether students performance in math is increased after recruiting two high qualified math teachers. Both pre-test and post-test scores of a random sample of 20 students were recorded. The mean difference between post-test and pre-test score was 15.4 with standard deviation 3.4. Does the data provide evidence to conclude that students performance in math is increased. Use 5% level of significance. Hypotheses and level of significance:
The hypotheses to be tested are:
Null hypothesis:
$H_0:{\mu }_d=0.$
Alternative hypothesis
$H_1:{\mu }_d>0.$
Assume the level of significance is given as $\alpha =0.05.$
Paired t-test statistic:
In order to test a hypothesis regarding whether the difference between a paired set of observations $(x_i,y_i)$ is significant or not, the paired t-test is used.
Test statistic:
The test statistic for the paired t-test is given as follows:
$t=\frac{\bar{d}}{se(\bar{d})}$
where,$\bar{d}=\frac{1}{n}\sum _{i=l}^n(x_i-y_i)$,
$se(\bar{d})=$ standard error of bard
$=\frac{standarderrorof\bar{d}}{n}n$
$=\sqrt{\frac{\frac{1}{n-1}\sum _{i=l}^n(d_i-\bar{d}{)}^2}{n}}$
The following table shows the necessary calculations for finding the test statistic:
$se(\bar{d})=\sqrt{\frac{\frac{1}{n-1}\sum _{i=l}^n(d_i-\bar{d}{)}^2}{n}}$
$=\frac{3.4}{\sqrt{20}}$
$\approx 0.7603$
$\therefore t=\frac{\bar{d}}{se(\bar{d})}$
$=\frac{15.4}{0.7603}$
$df=n-1$
$=20-1$
$=19$
Thus, $t=20.2552.$
Decision rule:
If the p-value is less than the level of significance then reject the null hypothesis. Otherwise fail to reject the null hypothesis
The p-value is given by:
The p-value $=0$
(From exel