Question

In: Statistics and Probability

Question 2: Related-Samples t-Test Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1...

Question 2: Related-Samples t-Test

Paired Samples Statistics

Mean

N

Std. Deviation

Std. Error Mean

Pair 1

freshman yr science score

51.8500

200

9.90089

.70010

senior yr science score

53.0450

200

12.10145

.85570

Paired Samples Correlations

N

Correlation

Sig.

Pair 1

freshman yr science score & senior yr science score

200

.878

.000

Paired Samples Test

Paired Differences

t

df

Sig. (2-tailed)

Mean

Std. Deviation

Std. Error Mean

95% Confidence Interval of the Difference

Lower

Upper

Pair 1

freshman yr science score - senior yr science score

-1.19500

5.83241

.41241

-2.00826

-.38174

-2.898

199

.004

  1. What was the dependent variable in this analysis?
  2. When did they test the students on that dependent variable?
  3. Describe the sample (using N, M, & SD for each time point).
  4. What are the null and alternative hypotheses for this study?
  1. Write out the t-test formula using the numbers given in the output. Make sure to write out the original formula in symbols before putting in the numbers from the output. You can either write out the formulas by hand and scan in the assignment or use the Equation Editor in Microsoft Word.
  2. Complete Hypothesis Test Step 4 by completing the following items:
  1. Decision about null hypothesis?
  2. Is it significant?
  3. Sentence in APA style.

Solutions

Expert Solution

Solution :-

(a):-

From given analysis, Science score is the Dependent variable.

(b):-

During their freshman and senior years in the school, They test the students on the dependent variable (i.e, science score).

(c):-

The example included N = 200 students that were tested in science in their Freshman recruit and senior years in school.

Freshman science score:

Mean,M = 51.8500.

Standard deviation,SD =  9.90089

Standard error of the mean = 0.70010

Senior year science score:

Mean,M = 53.0450.

Standard deviation, SD = 12.10145

Standard error of the mean = 0.85570.

(d):-

The null hypothesis for this examination was that there will be no significant contrast in 200 students science scores between their Freshman and senior years in school.

  • This can be written as .

The alternative hypothesis for this examination was that there will be a significant contrast in 200 students science scores between their Freshman and senior years in school.

  • This can be written as .

(a):-  Write out the t-test formula using the numbers given in the output?

Here we know the formula, i.e,

From given data,

i.e,  

(b):-  Complete Hypothesis Test Step 4 by completing the following items:

(i):-  Decision about null hypothesis?

Since the estimation of ±1.980 does not exceed the estimation of - 2.898, Hence we can reject the null hypothesis.

Additionally, since the significance value (2 - tailed) of 0.004 is less than the alpha of 0.05, Hence, the null hypothesis is rejected.

(ii):-  Is it significant?

Yes, this is significant that there is a sufficient evidence for science scores between 200 participants and the population

(ii):-  Sentence in APA style.

This results indicate that there is a significant difference in science scores between 200 participants and the population.


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