Respuesta :
Answer:
[tex]52.1-1.984\frac{12.3}{\sqrt{100}}=49.66[/tex]
[tex]52.1 +1.984\frac{12.3}{\sqrt{100}}=54.54[/tex]
The 95% confidence interval would be given by (49.66;54.54)
Step-by-step explanation:
Information given
[tex]\bar X= 52.1[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=12.3 represent the sample standard deviation
n=100 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=100-1=99[/tex]
The Confidence is 0.95 or 95%, and the significance would be [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case is: [tex]t_{\alpha/2}=1.984[/tex]
Replacing the info given we got:
[tex]52.1- 1.984\frac{12.3}{\sqrt{100}}=49.66[/tex]
[tex]52.1 +1.984\frac{12.3}{\sqrt{100}}=54.54[/tex]
The 95% confidence interval would be given by (49.66;54.54)
