Hypothesis testing: mean comparison

Question

I had great difficulties with an exam question and was wondering if any of you could help me understand it.

I was requested to do the following test at 5% level

$\left\{\begin{matrix}H_{0}:\mu _{1}=\mu _{2} \\ H_{1}:\mu _{1}\neq\mu _{2} \end{matrix}\right.$

I had the following data:

$\begin{matrix} n_{1}=11 & \bar{X_{1}}=7.25 & S_{1}^{2}=1.5\\ n_{2}=9 & \bar{X_{2}}=6.62 & S_{2}^{2}=1.2 \end{matrix}$

I guess I had to compute the following test statistic:

$T(\underline{X})=\frac{\bar{X_{1}}-\bar{X_{2}}-c}{\sqrt{\frac{n_{1}S_{1}^{2}+n_{2}S_{2}^{2}}{n_{1}+n_{2}-2}\frac{n_{1}+n_{2}}{n_{1}n_{2}}}}$

Where $c=0$ and $T(\underline{X})\sim t_{n_{1}+n_{2}-2}$

However, I was only given the standard normal distribution table and the Fisher distribution table.

I know that

$t_{n-1}:=\tfrac{N(0,1)}{\sqrt{\frac{\chi _{n-1}^{2}}{n-1}}}$

So I guess that (unsure):

$t_{n_{1}+n_{2}-2}:=\tfrac{N(0,1)}{\sqrt{\frac{\chi _{n_{1}+n_{2}-2}^{2}}{n_{1}+n_{2}-2}}}$

However, I don't have the $\chi^{2}$ distribution table. How can I still come up with a sensible answer?

Best regards

Answer 1

From the very sketchy description, it seems that this is a 2-sample t-test. I will do the 'pooled' version of this test, which assumes population variances are equal. That seems the closest match to the formula you give in your question. Output from Minitab 17 follows:

Sample   N  Mean  StDev  SE Mean
1       11  7.25   1.23     0.37
2        9  6.22   1.10     0.36

Difference = μ (1) - μ (2)
Estimate for difference:  1.030
95% CI for difference:  (-0.074, 2.134)
T-Test of difference = 0 (vs ≠): 
    T-Value = 1.96  P-Value = 0.066  DF = 18
Both use Pooled StDev = 1.1690

The pooled SD is $S_p = \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}} = 1.169.$ Then $T = \frac{\bar X_1 - \bar X_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}.$

I cannot account for the absence of a t table with your exam. Without such a table, two reasonable approaches are possible:

(i) The t statistic is $T = 1.96.$ For $n_1 + n_2 - 2 = 18$ degrees of freedom the t statistic is 'close' to standard normal but not exactly. If it were normal, the critical value would be to reject at the 5% level, if $|T| > 1.96.$ Your t value happens to be exactly 1.96. The critical value for $\mathsf{T}(18)$ has to be larger than 1.96. (All t distributions have heavier tails than standard normal.)

So you can guess that the sample means are not significantly different. (The exact 5% critical value for $\mathsf{T}(18)$ is $2.10,$ which confirms the statement above.) From R statistical software:

 qt(.975,18)
 ## 2.100922

(ii) You say you have a table of the F-distribution. You could use the fact that $T^2 \sim \mathsf{F}(1,18).$ From the F-table, you can see that he 5% critical value for $T^2$ is $4.41.$ Therefore, you cannot reject at the 5% level (no guessing involved) because $T^2 = 3.84 < 4.41.$ Notice that (using either t table or F table) the P-value is $P(|T| > 1.96) = P(T^2 > 3.84) = 0.066 > .05.$

qf(.95, 1, 18)
## 4.413873

2*pt(-1.96, 18)
## 0.06566396
1-pf(1.96^2, 1, 18)
## 0.06566396

The figure below shows PDFs of $\mathsf{T}(18)$ [left] and $\mathsf{F}(1,18)$ with critical values as vertical red lines and observed values of $T$ and $F$ as vertical black lines.