Question

Two new mathematics learning techniques are being tested. The researchers are unsure which one, if any, will be better. Two hundred students were randomly selected from a population. Ninety of them were randomly assigned to use Technique A, and 110 of them were randomly assigned to use Technique BEach student spent 30 minutes learning the technique to which they were assigned, and then were asked to complete a task. The time to complete the task was recorded, in seconds. A shorter time indicates better mastery ofthe task. The data are below:Technique A: sample average = 26.1, sample SD = 7.2Technique B: sample average = 27.9, sample SD = 12.4(a) State hypotheses relevant to the research question.(b) Perform a test of the hypotheses from (a) using a significance level of 10%. Make sure to compute the test statistic and P-value, and make a conclusion in context.

Answer 1

Answer:

(a) Hypotheses relevant to the research question are  

$H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} < 0$ (lower-tail alternative) and

$H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).  

(b) There is no evidence that one mathematic learning technique is better than the other.

Step-by-step explanation:

Let's suppose that the data related to Technique A comes from population 1 and that the data related to Technique B comes from population 2. We have large sample sizes $n_{1} = 90$  and $n_{2} = 110$. The unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$, i.e., 26.1 - 27.9 = -1.8. The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}\approx$ $\sqrt{\frac{(7.2)^{2}}{90}+\frac{(12.4)^{2}}{110}}$ = 1.4049.

(a) Hypotheses relevant to the research question are  

$H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} < 0$ (lower-tail alternative) and

$H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).  

(b) The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{-1.8}{1.4049} = -1.2812$.  

The p-value for the lower-tail alternative is P(Z < -1.2812) =  0.1000617 and the p-value for the upper-tail alternative is P(Z > -1.2812) = 0.8999. The p-value is greater than 0.10 in both cases, therefore, we fail to reject the null hypotheses in both cases.