Question

Use student-t to derive the following paired test
set.seed(20)
girls_height<- rnorm(50, mean = 160, sd = 30)
boys_height<- rnorm(50, mean = 172, sd = 30)

#-------------------------------------------------------------
State the hypothesis for the paired-test of the mean of
# girls' height and boys' height.

Check the three assumptions before running the student t test
# for the samples. You need to write and run the code, and
# report the results.

Answer 1

The required commands generate the required variables. The null hypothesis of the paired t-test of the mean of two groups would be and the alternate would be , where D is the true/population mean difference.

The assumptions and their verification are as below (note that there are different assumptions in different text books, but major four assumptions are stated).

• The variables must be interval or ratio scale. This can be checked by looking at the data itself, via simple R-commands as below.
  ----------------------------------------------------
  > View(girls_height)
  > View(boys_height)
  ----------------------------------------------------
• The observations are independent from one another. The test of correlation coefficient can be done in this regard as below. The null would be , and alternate would be . The t-statistic would be .
  ----------------------------------------------------

  > cor.test(girls_height,boys_height)

     Pearson's product-moment correlation

  data: girls_height and boys_height
  t = 1.2589, df = 48, p-value = 0.2142
  alternative hypothesis: true correlation is not equal to 0
  95 percent confidence interval:
  -0.1047890 0.4354522
  sample estimates:
        cor
  0.1787731

  ----------------------------------------------------
  As can be seen, p-value is 0.2142, meaning that the test statistic is not significant at 1% or 5%, or even 10%. Hence, the two variables are independent of each other. The critical t at 10% alpha would be , and the calculated t is less than that.
• The variables should be normally distributed. This can be checked via Jarque-Bera test (JB test), in which the null hypothesis is that the variable is not different from a normal distribution, and the alternate is that the variable is indeed different from a normal distribution. The R-commands are as below (package 'tseries' is required, along with dependencies of that).
  ----------------------------------------------------

  > jarque.bera.test(girls_height)

     Jarque Bera Test

  data: girls_height
  X-squared = 0.68004, df = 2, p-value = 0.7118

  > jarque.bera.test(boys_height)

     Jarque Bera Test

  data: boys_height
  X-squared = 0.24858, df = 2, p-value = 0.8831

  ----------------------------------------------------
  As can be seen, both the p-values fail to reject the null, which means that both the variables are not significantly different from a normal distribution.
• The variables should not contain any outlier. This can be confirmed via Grubb's test, with the R-commands as below (this requires the package 'outliers', may be installed without dependencies).
  ----------------------------------------------------

  > grubbs.test(girls_height)

     Grubbs test for one outlier

  data: girls_height
  G = 2.61871, U = 0.85719, p-value = 0.1704
  alternative hypothesis: lowest value 73.3084715675764 is an outlier

  > grubbs.test(boys_height)

     Grubbs test for one outlier

  data: boys_height
  G = 2.33549, U = 0.88641, p-value = 0.4174
  alternative hypothesis: lowest value 112.95922652111 is an outlier

  ----------------------------------------------------
  As can be seen, both the p-values fail to reject the null that there is no outliers.

As all the assumptions are verified, we may proceed to do the paired t-test. This can be done manually and via R-commands. Both are done as below.

The commands to test the variables directly is as below.

----------------------------------------------------

> t.test(girls_height,boys_height,paired = T)

   Paired t-test

data: girls_height and boys_height
t = -4.0983, df = 49, p-value = 0.0001559
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-32.60696 -11.15059
sample estimates:
mean of the differences
              -21.87877

----------------------------------------------------

As can be seen, the mean difference is significantly different from zero since the t-statistic is significant at 5% or even at 1% alpha. The low p-value of 0.0001559, less than 0.05 or 0.01 states that we may reject the null hypothesis, which states true mean is equal to zero.

The manual method would be as below.

----------------------------------------------------

> d <- girls_height - boys_height
> t <- mean(d)/(sd(d)/(50^0.5))
> abs(t) > qt(0.995,50-1)
[1] TRUE
> (pt(t,50-1))*2
[1] 0.0001559402

----------------------------------------------------

The difference variable is basically , and the t-statistic is , and since the standard error of the difference variable would be , for sd(d) be the sample (not population) standard deviation, we have the t-statistic as . The founded t-statistic is also equal to the direct command t-statistic. As the absolute value of the t-statistic is greater than the critical t at 0.01 alpha, we reject the null that the true difference is zero. The the critical t is . The p-value is also given just as before, which is less than 0.01 alpha.

Hence, the conclusion would be that, the difference between the height of girls and boys for the given data is significantly different.