You want to rent an unfurnishedone-bedroom
apartment for next semester. The mean monthlyrent for a random
sample of 10 apartments advertised in the localnewspaper is $580.
Assume that the standard deviation is$90. Find a 95% confidence
interval for the mean monthly rentfor unfurnished one bedroom
apartments available for rent in thiscommunity.

Respuesta :

Answer: [tex](\$524.22,\ \$635.78)[/tex]

Step-by-step explanation:

Confidence interval for population mean is given by :-

[tex]\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\overline{x}[/tex] = Sample mean

z* =  critical z-value.

[tex]\sigma[/tex] = Population standard deviation.

n= Sample size.

Let x be the denotes the monthly rent for unfurnished one bedroom  apartments available for rent in this community.

As per given , we have

n= 10

[tex]\overline{x}=\$580[/tex]

[tex]\sigma=\$90[/tex]

Critical value for 95% confidence : z* = 1.96

So the 95% confidence  interval becomes,

[tex]580\pm (1.96) \dfrac{90}{\sqrt{10}}[/tex]

[tex]=580\pm (1.96) \dfrac{90}{3.162278}[/tex]

[tex]=580\pm (1.96)(28.46)[/tex]

[tex]=580\pm 55.78[/tex]

[tex]=(580-55.78,\ 580+55.78)=(524.22,\ 635.78)[/tex]

Hence, a 95% confidence  interval for the mean monthly rent for unfurnished one bedroom  apartments available for rent in this community= [tex](\$524.22,\ \$635.78)[/tex]