Research Methods
1. Assume the room rate for a 3-star hotel in Paris follows a normal distribution with a mean of 175€ and a standard deviation of 16€. What is the probability that a randomly selected 3-star hotel has a room rate of
Z(X) = (x-μ)/σ σ=16 μ=175
a) Less than 195 Euros?
Z (195) = (195-175)/16 = 1, 25 so 89, 4%
b) More than 180 Euros?
Z (180) = (180-175)/16 = 0, 31 so 62, 2%
c) Exactly 170 Euros?
Z (170) = (170-175)/16 = - 0, 31 so 37, 8%
d) Between 150 and 182 Euros?
Z (150) = (150-175)/16 = - 1, 56
Z (182) = (182-175)/16 = 0, 44
P (150 < X < 182) = - 1, 56 – 0, 44 = - 2 so 2, 3%
2. A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 55 families will be between 17 and 18 pounds.
μ=17,2 σ=2,5 n=55 σ(X)= σ/(√n) Z(X) = (x ̅-μ(x))/(σ(x))
σ(X)= (2,5)/√55=0,34 Z(17)= (17-17,2)/(0,34)= -0,59 Z(18)= (18-17,2)/(0,34)=2,37
P (17 < X < 18) = - 0, 59 – 2, 37 = - 2, 96 so 0, 15%
3. According to the government-lending institution Sallie Mae, students graduating from college have an average credit card debt of $4,100. A random sample of 50 graduating seniors was selected, and their average credit card debt was found to be $4,360. Assume the standard deviation for student credit card debt is $1,200. Using , answer the following questions.
a) Does this sample provide enough evidence to challenge the findings by Sallie Mae?
N= 50 α=0,05 σ=1 200 x ̅=4 360
This is a two-tails test: 〖 H〗_0: μ=4100 df = n – 1 = 49
〖 H〗_1: μ ≠4100
Z (x ̅)= (x ̅- μ_(H_0 ))/(σ/(√n)) = (4360-4100)/(1200/(√50)) = 1, 53
Here Z (x ̅)=1,53 α so we don’t reject H_0, the P. value confirmed the precedent result.
4. Low profile tires are
Z(X) = (x-μ)/σ σ=16 μ=175
a) Less than 195 Euros?
Z (195) = (195-175)/16 = 1, 25 so 89, 4%
b) More than 180 Euros?
Z (180) = (180-175)/16 = 0, 31 so 62, 2%
c) Exactly 170 Euros?
Z (170) = (170-175)/16 = - 0, 31 so 37, 8%
d) Between 150 and 182 Euros?
Z (150) = (150-175)/16 = - 1, 56
Z (182) = (182-175)/16 = 0, 44
P (150 < X < 182) = - 1, 56 – 0, 44 = - 2 so 2, 3%
2. A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 55 families will be between 17 and 18 pounds.
μ=17,2 σ=2,5 n=55 σ(X)= σ/(√n) Z(X) = (x ̅-μ(x))/(σ(x))
σ(X)= (2,5)/√55=0,34 Z(17)= (17-17,2)/(0,34)= -0,59 Z(18)= (18-17,2)/(0,34)=2,37
P (17 < X < 18) = - 0, 59 – 2, 37 = - 2, 96 so 0, 15%
3. According to the government-lending institution Sallie Mae, students graduating from college have an average credit card debt of $4,100. A random sample of 50 graduating seniors was selected, and their average credit card debt was found to be $4,360. Assume the standard deviation for student credit card debt is $1,200. Using , answer the following questions.
a) Does this sample provide enough evidence to challenge the findings by Sallie Mae?
N= 50 α=0,05 σ=1 200 x ̅=4 360
This is a two-tails test: 〖 H〗_0: μ=4100 df = n – 1 = 49
〖 H〗_1: μ ≠4100
Z (x ̅)= (x ̅- μ_(H_0 ))/(σ/(√n)) = (4360-4100)/(1200/(√50)) = 1, 53
Here Z (x ̅)=1,53 α so we don’t reject H_0, the P. value confirmed the precedent result.
4. Low profile tires are