Question

Suppose we want to test whether there is a difference between the room rates of luxury hotels in New York versus Los Angeles. We collect data from 20 random luxury hotels in New York and 30 random luxury hotels in L.A. The standard deviation of the data from New York is $40 and the standard deviation of the data from L.A. is $50. What is the pooled-variance standard error for a test or confidence interval for the difference in the means? The standard error is

Round your answer to two decimal places.

Answer 1

The pooled-variance standard error calculated for the difference in room rates of luxury hotels between New York and Los Angeles is approximately $12.78 when rounded to two decimal places.

Step-by-step explanation:

To calculate the pooled-variance standard error for the difference in means between two independent samples, we use the following formula:

SE = √[(s1² / n1) + (s2² / n2)]

Where:

s1 and s2 are the standard deviations of the two samples.

n1 and n2 are the sample sizes of the two samples.

For the luxury hotels in New York and Los Angeles, we have the following values:

• s1 = $40 (Standard deviation for New York)
• n1 = 20 (Sample size for New York)
• s2 = $50 (Standard deviation for Los Angeles)
• n2 = 30 (Sample size for Los Angeles)

The pooled-variance standard error can be calculated as follows:

SE = √[($40² / 20) + ($50² / 30)]

SE = √[(1600 / 20) + (2500 / 30)]

SE = √[80 + 83.33]

SE = √[163.33]

SE ≈ $12.78 (rounded to two decimal places)

This is the standard error you would use when performing a hypothesis test or constructing a confidence interval for the difference between the means of room rates for luxury hotels in New York and Los Angeles.

Answer 2

the answer is $335.16.

Step-by-step explanation:

The formula for the pooled variance standard error is:

s_pooled = sqrt(((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2))

where n1 and n2 are the sample sizes, s1 and s2 are the sample standard deviations.

Substituting the given values, we get:

s_pooled = sqrt(((20 - 1)(40)^2 + (30 - 1)(50)^2)/(20 + 30 - 2))

s_pooled = sqrt(((19)(1600) + (29)(2500))/48)

s_pooled = sqrt(112316.67)

s_pooled = 335.16

Therefore, the pooled-variance standard error is $335.16. Rounded to two decimal places, the answer is $335.16.