The crisis in the housing market during the most recent economic recession led to lower vacancy rates in the apartment industry. This, in turn, has resulted in an increase in monthly rents for apartments. Suppose a random sample of 40 apartments had an average rent of $1,090. Assume the standard deviation for the population of apartment rents is $260. Determine the 98% confidence interval for this sample.

(994.37, 1185.62)

Step-by-step explanation:

Given that :

Mean (m) = 1090

Standard deviation (s) = 260

Samole size (n) = 40

α = 98%

Using the relation :

Confidence interval = mean ± Error

Error = Zcritical * (standard deviation / sqrt (n))

Zcritical at 98% = 2.326

Error = 2.326 * (260 / sqrt(40))

Error = 95.620951

Hence,

Confidence interval :

Lower boundary = 1090 - 95.620951 = 994.379049

Upper boundary = 1090 + 95.620951 = 1185.620951

(994.37, 1185.62)

the answer should be:

a. 3, 4, and 6

b. 18, 24, and 36

c. 4.8, 6.4, and 9.6

hope i

-lavira : )