Respuesta :
Answer:
[tex] df = n-1=20-1=19[/tex]
The confidence level for this case is 90% so then the significance level is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] so then the critical value in the t distribution with df =19 is given by :
[tex] t_{\alpha/2}= 1.729[/tex]
Now we can find the margin of error for the confidence interval given by:
[tex] ME = t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
And replacing we got:
[tex] ME = 1.729 *\frac{4.2}{\sqrt{20}}= 1.624[/tex]
And the confidence interval would be given by:
[tex] 43.7 \pm 1.62[/tex]
And the best option is:
a. 43.7 + 1.62.
Step-by-step explanation:
The random variable of interest for this problem is the score for all marathon runners X and in special we want to infer about the true mean [tex]\mu[/tex]. We have the following info given:
[tex]\bar X= 43.7[/tex] the sample mean for the test scores
[tex] s= 4.2[/tex] the sample deviation given
[tex]n = 20[/tex] the sample size selected
And we want to construct a confidence interval for the true mean given by this formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The degrees of freedom for this case are given by:
[tex] df = n-1=20-1=19[/tex]
The confidence level for this case is 90% so then the significance level is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] so then the critical value in the t distribution with df =19 is given by :
[tex] t_{\alpha/2}= 1.729[/tex]
Now we can find the margin of error for the confidence interval given by:
[tex] ME = t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
And replacing we got:
[tex] ME = 1.729 *\frac{4.2}{\sqrt{20}}= 1.624[/tex]
And the confidence interval would be given by:
[tex] 43.7 \pm 1.62[/tex]
And the best option is:
a. 43.7 + 1.62.
