Testing A Hypothesis

Agent Smith

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Oct 18, 2023
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4 Spanish teachers take an intensive summer course in Spanish. Their pre-course and post-course scores in Spanish are given below:
SubjectPre-course scorePost-course score
13029
22630
33035
41518
The course organizers hope to show that taking their course improves Spanish for the subjects.
Formulate a null hypothesis and an alternative hypothesis and test them.
Significance level α=0.10\alpha = 0.10

Construct a 90% confidence interval for the mean increase in scores.

Computations I did
:
Mean pre-course score = μ1=25.25\mu_1 = 25.25
Standard deviation for pre-course score = σ1=6.14\sigma_1 = 6.14

Mean post-course score = μ2=28\mu_2 = 28
Standard deviation for post-course score = σ2=6.2\sigma_2 = 6.2

What I did:
For μ2=28\mu_2 = 28, z score = 2825.256.14\frac{28 - 25.25}{6.14}. Now that I think of it, this is wrong and so I won't bother posting the rest of what I did wrong.

What I should've done:
μ2μ1=μD\mu_2 - \mu_1 = \mu_D, the mean difference between pre-course and post-course scores
μD=2.75\mu_D = 2.75
σD2=σ12+σ22=6.142+6.22    σD=8.73\sigma_D ^2 = \sigma_1 ^2 + \sigma_2 ^2 = 6.14^2 + 6.2^2 \implies \sigma_D = 8.73

Now to formulate the hypotheses:
Assumption: If there's no difference the post-course distribution of scores should be identical to the pre-course distribution of scores and so:
H0:μD=μ1μ1=0H_0: \mu_D = \mu_1 - \mu_1 = 0.
The standard deviation for the difference of means = σ1\sigma_1.

Ha:μD>0H_a: \mu_D > 0

z score for a difference in mean scores of 2.752.75 is given by 2.7506.14=0.45\frac{2.75 - 0}{6.14} = 0.45
P-value associated with z score = 0.450.45 is 10.6736=0.32641 - 0.6736 = 0.3264
P-value > α\alpha and so we can't reject H0H_0.

I'm not sure I got this right.


??? 🤔

Constructing the 90% confidence interval for the mean increase in scores:
For a 90% confidence interval (CI), we need the z-score for 0.95 probability, which is 1.64
So the 90% CI is μD±z×σD=2.75±(1.64×8.73)=2.75±14.32\mu_D \pm z^* \times \sigma_D = 2.75 \pm (1.64 \times 8.73) = 2.75 \pm 14.32
The 90% CI is [11.57,17.07][-11.57, 17.07]

The computed 90% CI seems to square with my conclusion that I can't reject H0H_0 as 00 lies between 11.57-11.57 and 17.0717.07. It is plausible that there's no difference between pre-course and post-course test. In addition, there's even a chance that the course impacts negatively on Spanish language skills (11.57<0-11.57 < 0

Sorry for the long post, but I really need help on this. Is this correct?
 
Are the two samples, Pre and Post\text{Pre and Post}, paired or independent?

I see them paired. What about you?
 
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