# Testing $H_0 : \mu_x \neq \mu_y$, in a company that markets two brands of latex paint.

A company markets two brands of latex paint regular and a more expensive brand that claims to dry an hour faster. A consumer magazine decides to test this claim by painting ten panels with each product. The average drying time of the regular brand is $2.1$ hours with a sample standard deviation of $12$ minutes. The fast-drying version has an average of $1.6$ hours with a sample standard deviation of $16$ minutes. Test the null hypothesis that the more expensive brand dries an hour quicker. Use one sided $H_1$. Let $\alpha = 0.05$

attempt: Given $n = 10$, $m = 10$, $\bar x = 2.1, \bar y = 1.6, S_x = 12, S_Y = 16$.

To test $H_0 : \mu_x \neq \mu_y$ vs. $H_1 : \mu_x < \mu_y$, reject $H_0$ if $t \leq - t_{\alpha, n+m-2}$.

And $S_p = \sqrt{\frac{(n-1)S_x^2 + (m-1)S_Y^2}{(n+m-2)}} = \sqrt{\frac{(10-1)12^2 + (10-1)16^2}{(10 + 10 -2)}} = 14.9$

Then $t = \frac{\bar x - \bar y}{S_p \sqrt(\frac{1}{n} + \frac{1}{m})} = \frac{2.1 - 1.6}{14.9 \sqrt(\frac{1}{10} + \frac{1}{10})} = 0.0750$

And $t_{\alpha, n+m-2} = t_{0.05/2, 10 + 10-2} = t_{0.025, 18} = 2.1009$

So since$t \leq - t_{\alpha, n+m-2}$, we reject $H_0$.

Is this correct? Can anyone please help? I don't now if I have set up the problem correctly. Thank you in advance.

## 2 Answers

The first problem you have is that you need to set up the hypothesis correctly. In your case, because you are asked to test that the "fast" paint is faster by at least an hour, the structure of this hypothesis is a superiority by a margin test.

The null hypothesis is that there is no difference in the mean drying times: $$H_0 : \mu_r - \mu_f < \epsilon,$$ where $r$ denotes "regular" and $f$ denotes "fast," and $\epsilon = 60$ is the margin of superiority in minutes. But the requirement we are given in order to reject $H_0$ is that the drying time of the fast paint is at least an hour faster, so this is $$H_a : \mu_r - \mu_f \ge \epsilon.$$

Under the assumption that the null hypothesis is true, the distribution of the test statistic $$T = \frac{\bar r - \bar f - \epsilon}{\sqrt{\frac{s_{\vphantom f r}^2}{n_r} + \frac{s_f^2}{n_f}}} \sim t_{\nu}$$ is Student's $t$ distributed with $$\nu \approx \frac{\left(\frac{s_{\vphantom f r}^2}{n_r} + \frac{s_f^2}{n_f}\right)^{\!2}}{\frac{s_{\vphantom f r}^4}{n_r^2 (n_r-1)} + \frac{s_f^4}{n_f^2 (n_f-1)}}$$ degrees of freedom. In your case, $n_f = n_r = n = 10$, so the above formulas reduce to $$T = \frac{\bar r - \bar f - \epsilon}{\sqrt{\frac{s_{\vphantom f r}^2 + s_f^2}{n}}}, \quad \nu \approx \frac{(n - 1)(s_{\vphantom f r}^2 + s_f^2)^2}{s_{\vphantom f r}^4 + s_f^4 }.$$ Now with $\bar f = 96$, $\bar r = 126$, $s_f = 16$, $s_r = 12$, we obtain the test statistic $$T = -4.74342, \quad \nu \approx 16.69.$$ But since $T < 0$ the test automatically fails to reject $H_0$ (the critical value is on the upper tail of the distribution, but the calculated statistic is on the lower tail). So we immediately conclude that there is insufficient evidence to suggest that the fast-drying paint dries at least an hour faster than the regular paint: but this much should have been obvious by simply comparing the difference in sample means and noting that $126$ minutes is not more than $60$ minutes greater than $96$ minutes. In light of this, it seems odd to ask such a question; perhaps you might want to double check the statement of the problem to be sure it is faithful.

First, you have to put everything into minutes.

If you do that, you should get that $S_p = 14.14$ (the difference from your answer might be that you rounded prematurely).

Then, for the main difference: I get $T = 4.74$, which certainly leads to rejection. (With your incorrect $T = 0.075,$ you would certainly not reject; you seem to have some misunderstanding how to use critical values from tables.)

Note: I would tend to use a Welch separate-variances t test instead of a pooled t test. However, (1) I don't know whether you've studied that, and (2) you'll reject with either test.

For a pooled test, $DF = 10 + 10 - 2 = 18.$ The Welch test has a more complicated formula for DF, and the result is $DF = 16.$ Also, because the two sample sizes are equal, pooled and Welch have the same value of the $T$ statistic. So it doesn't really matter which you use.