Respuesta :
Answer:
[tex] (75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96[/tex]
[tex] (75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96[/tex]
And the confidence interval would be given by:
[tex] -1.96 \leq \mu_1 -\mu_2 \leq 7.96[/tex]
And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.
Step-by-step explanation:
For this case we have the following info given :
[tex] \bar X_1= 75.1[/tex] represent the sample mean for the scores of the undergraduate students
[tex]s_1 = 12.8[/tex] represent the standard deviation for the undergraduate students
[tex]n_1 =35[/tex] the sample size for the undergraduate
[tex] \bar X_2= 72.1[/tex] represent the sample mean for the scores of the high school students
[tex]s_2 = 14.6[/tex] represent the standard deviation for the high school students
[tex]n_2 =50[/tex] the sample size for the high school
The confidence interval for the true difference of means is given by:
[tex] (\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}[/tex]
The degrees of freedom are given by:
[tex] df=n_1 +n_2 -2= 35+50-2=83[/tex]
The confidence level is 90% and the significance level is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] then the critical value would be:
[tex]t_{\alpha/2}= 1.99[/tex]
And replacing the info we got:
[tex] (75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96[/tex]
[tex] (75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96[/tex]
And the confidence interval would be given by:
[tex] -1.96 \leq \mu_1 -\mu_2 \leq 7.96[/tex]
And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.
