Question

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

Answer 1

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

Part b) Interval: `(-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: mu_"y" = mu_"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 `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 >= 10*14`
• in 2016 is greater than 200 `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{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`

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

`"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.

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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:

`(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) +- (2.160)sqrt(frac{3.2^2}{14}+frac{4.6^2}{20})`

`2.2 +- 2.889`

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