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

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

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Answer 1 · 2018-05-24T03:44:20

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

Part b) Interval: $(- 0.689, 5.089)$ $(-0.689, 5.089)$

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:

$H_{0} : μ_{y} = μ_{x}$ $H_0: mu_"y" = mu_"x"$
$H_{a} : μ_{y} < μ_{x}$ $H_a: mu_"y" < mu_"x"$

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 $α = .05$ $alpha = .05$ 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 $N_{x} \geq 10 \cdot 14$ $N_x >= 10*14$
in 2016 is greater than 200 $N_{y} \geq 10 \cdot 20$ $N_y >= 10*20$
Large/Normal samples: We must assume that the distributions for both populations are each approximately Normal

The formula for the test statistic t is:

$t = \frac{¯ x - ¯ y}{\sqrt{\frac{s_{x}^{2}}{n_{x}} + \frac{s_{y}^{2}}{n_{y}}}}$ $t = frac{barx - bary}{sqrt(frac{s_x^2}{n_x}+frac{s_y^2}{n_y})}$

with degrees of freedom (using the lower n) $df = n_{x} - 1$ $"df" = n_x-1$

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

$df = 14 - 1 = 13$ $"df" = 14-1=13$

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

$p = .0549$ $p = .0549$

Since our $p$ $p$ value $p = .0549$ $p = .0549$ is greater than our significance level $α = .05$ $alpha = .05$ , 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:

$(¯ x - ¯ y) \pm t^{*} \sqrt{\frac{s_{x}^{2}}{n_{x}} + \frac{s_{y}^{2}}{n_{y}}}$ $(barx - bary) +- t^("*")sqrt(frac{s_x^2}{n_x}+frac{s_y^2}{n_y})$

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

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

$2.2 \pm 2.889$ $2.2 +- 2.889$

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

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

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