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:
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 [imath]\alpha = 0.10[/imath]
Construct a 90% confidence interval for the mean increase in scores.
Computations I did:
Mean pre-course score = [imath]\mu_1 = 25.25[/imath]
Standard deviation for pre-course score = [imath]\sigma_1 = 6.14[/imath]
Mean post-course score = [imath]\mu_2 = 28[/imath]
Standard deviation for post-course score = [imath]\sigma_2 = 6.2[/imath]
What I did:
For [imath]\mu_2 = 28[/imath], z score = [imath]\frac{28 - 25.25}{6.14}[/imath]. 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:
[imath]\mu_2 - \mu_1 = \mu_D[/imath], the mean difference between pre-course and post-course scores
[imath]\mu_D = 2.75[/imath]
[imath]\sigma_D ^2 = \sigma_1 ^2 + \sigma_2 ^2 = 6.14^2 + 6.2^2 \implies \sigma_D = 8.73[/imath]
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:
[imath]H_0: \mu_D = \mu_1 - \mu_1 = 0[/imath].
The standard deviation for the difference of means = [imath]\sigma_1[/imath].
[imath]H_a: \mu_D > 0[/imath]
z score for a difference in mean scores of [imath]2.75[/imath] is given by [imath]\frac{2.75 - 0}{6.14} = 0.45[/imath]
P-value associated with z score = [imath]0.45[/imath] is [imath]1 - 0.6736 = 0.3264[/imath]
P-value > [imath]\alpha[/imath] and so we can't reject [imath]H_0[/imath].
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 [imath]\mu_D \pm z^* \times \sigma_D = 2.75 \pm (1.64 \times 8.73) = 2.75 \pm 14.32[/imath]
The 90% CI is [imath][-11.57, 17.07][/imath]
The computed 90% CI seems to square with my conclusion that I can't reject [imath]H_0[/imath] as [imath]0[/imath] lies between [imath]-11.57[/imath] and [imath]17.07[/imath]. 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 ([imath]-11.57 < 0[/imath]
Sorry for the long post, but I really need help on this. Is this correct?
| Subject | Pre-course score | Post-course score |
| 1 | 30 | 29 |
| 2 | 26 | 30 |
| 3 | 30 | 35 |
| 4 | 15 | 18 |
Formulate a null hypothesis and an alternative hypothesis and test them.
Significance level [imath]\alpha = 0.10[/imath]
Construct a 90% confidence interval for the mean increase in scores.
Computations I did:
Mean pre-course score = [imath]\mu_1 = 25.25[/imath]
Standard deviation for pre-course score = [imath]\sigma_1 = 6.14[/imath]
Mean post-course score = [imath]\mu_2 = 28[/imath]
Standard deviation for post-course score = [imath]\sigma_2 = 6.2[/imath]
What I did:
For [imath]\mu_2 = 28[/imath], z score = [imath]\frac{28 - 25.25}{6.14}[/imath]. 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:
[imath]\mu_2 - \mu_1 = \mu_D[/imath], the mean difference between pre-course and post-course scores
[imath]\mu_D = 2.75[/imath]
[imath]\sigma_D ^2 = \sigma_1 ^2 + \sigma_2 ^2 = 6.14^2 + 6.2^2 \implies \sigma_D = 8.73[/imath]
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:
[imath]H_0: \mu_D = \mu_1 - \mu_1 = 0[/imath].
The standard deviation for the difference of means = [imath]\sigma_1[/imath].
[imath]H_a: \mu_D > 0[/imath]
z score for a difference in mean scores of [imath]2.75[/imath] is given by [imath]\frac{2.75 - 0}{6.14} = 0.45[/imath]
P-value associated with z score = [imath]0.45[/imath] is [imath]1 - 0.6736 = 0.3264[/imath]
P-value > [imath]\alpha[/imath] and so we can't reject [imath]H_0[/imath].
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 [imath]\mu_D \pm z^* \times \sigma_D = 2.75 \pm (1.64 \times 8.73) = 2.75 \pm 14.32[/imath]
The 90% CI is [imath][-11.57, 17.07][/imath]
The computed 90% CI seems to square with my conclusion that I can't reject [imath]H_0[/imath] as [imath]0[/imath] lies between [imath]-11.57[/imath] and [imath]17.07[/imath]. 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 ([imath]-11.57 < 0[/imath]
Sorry for the long post, but I really need help on this. Is this correct?