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

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1 Answer
May 24, 2018

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.


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.