# Are population means score in statistics different between 2006 and 2016 students? ( Help ) (Stats)

## May 24, 2018

Part a) No, there is not a significant difference between 2006 and 2016 students' test scores.

Part b) Interval: $\left(- 0.689 , 5.089\right)$

#### Explanation:

Part A

We want to determine whether there is a significant difference between the population mean scores of the students. We will test the hypotheses:

${H}_{0} : {\mu}_{\text{y" = mu_"x}}$
${H}_{a} : {\mu}_{\text{y" < mu_"x}}$

where "y" represents data from 2016, and "x" represents data from 2006.

We can conduct a two-sample t-test for the difference between two means using a significance level $\alpha = .05$ if the following conditions for inference are met.

• Random: both samples (2006 and 2016) are random samples
• 10% condition / Independence: We must assume that the population of test takers ...
• in 2006 is greater than 140 ${N}_{x} \ge 10 \cdot 14$
• in 2016 is greater than 200 ${N}_{y} \ge 10 \cdot 20$
• Large/Normal samples: We must assume that the distributions for both populations are each approximately Normal

The formula for the test statistic t is:

$t = \frac{\overline{x} - \overline{y}}{\sqrt{\frac{{s}_{x}^{2}}{{n}_{x}} + \frac{{s}_{y}^{2}}{{n}_{y}}}}$

with degrees of freedom (using the lower n) $\text{df} = {n}_{x} - 1$

Substitute values:
$t = \frac{73.0 - 70.8}{\sqrt{\frac{{3.2}^{2}}{14} + \frac{{4.6}^{2}}{20}}}$

$\text{df} = 14 - 1 = 13$

Now, using the table of t critical values or a calculator, we can find that our $p$ value lies between $.05$ and $.10$.
(Using a calculator):

$p = .0549$

Since our $p$ value $p = .0549$ is greater than our significance level $\alpha = .05$, we fail to reject our null hypothesis. There is not convincing evidence of a significant difference between 2006 and 2016 tests.

Part B

We want to construct a 95% confidence interval for the difference between two means.

We can use a two-sample t-interval . The conditions for inference were verified in part (a).

The formula for the two-sample t-interval with 95% confidence is:

$\left(\overline{x} - \overline{y}\right) \pm {t}^{\text{*}} \sqrt{\frac{{s}_{x}^{2}}{{n}_{x}} + \frac{{s}_{y}^{2}}{{n}_{y}}}$

Substitute values:
Find t using the table of critical t values (linked above). This is where we specify the 95% confidence.

$\left(73.0 - 70.8\right) \pm \left(2.160\right) \sqrt{\frac{{3.2}^{2}}{14} + \frac{{4.6}^{2}}{20}}$

$2.2 \pm 2.889$

We are 95% confident that the interval from $- 0.689$ to $5.089$ captures the true difference $\overline{x} - \overline{y}$ between the population mean test scores in 2006 and 2016.