Answer: The lower limit of this confidence interval = $576.41
The upper limit of this confidence interval = $831.09
Step-by-step explanation:
Let [tex]\mu[/tex] be the population mean price of a used road bike.
As per given have,
n=11
Degree of freedom = 10
[tex]\overline{x}=\$703.75[/tex]
s= $189.56
T-critical value for 95% confidence :
[tex]t_{(df, \alpha/2)}=t_{(10,0.025)}=2.228[/tex]
Now, 95% confidence interval for μ, the mean price of a used road bike. will be :
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
[tex]\$703.75\pm (2.228)\dfrac{189.56}{\sqrt{11}}[/tex]
[tex]\$703.75\pm \$127.34[/tex]
[tex](\$703.75-\$127.34,\ \$703.75+\$127.34)[/tex]
[tex](\$576.41,\ \$831.09)[/tex]
Thus , the lower limit of this confidence interval = $576.41
The upper limit of this confidence interval = $831.09
