Respuesta :
Answer:
[tex] (17-14) -1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=1.222[/tex]
[tex] (17-14) +1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=4.778[/tex]
So then we can conclude at 90% of confidence that the difference in the two means is between 1.222 and 4.778 beats per minute
Step-by-step explanation:
For this case we have the following info given:
[tex] \bar X_1 = 17[/tex] sample mean for the first sample
[tex] \bar X_2 = 14[/tex] sample mean for the second sample
[tex]\sigma =4.2[/tex] represent the population deviation
[tex] n_1= n_2 =30[/tex] represent the sample size ofr each case
We can construct the confidence interval for the difference of means with the following formula:
[tex] (\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}}[/tex]
And the confidence for this case is 90% or 0.9 so then the significance level is [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] the critical value for this case is:
[tex] z_{\alpha/2}=1.64[/tex]
And replacing we got:
[tex] (17-14) -1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=1.222[/tex]
[tex] (17-14) +1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=4.778[/tex]
So then we can conclude at 90% of confidence that the difference in the two means is between 1.222 and 4.778 beats per minute
