The California Chamber of Commerce has asked you to do a study concerning people
ID: 3172889 • Letter: T
Question
The California Chamber of Commerce has asked you to do a study concerning people's attitudes toward our city. As part of the study, you will ask them to rate their image of California on a scale from 1 to 100 (1= awful city - call in the bulldozers; 100 = wonderful city - there is none better). It is known that this scale is normally distributed with an average rating of 50 and a standard deviation of 12.
c. Suppose a sample of 36 people is chosen. Their sample mean is 54. Can you conclude that the attitude toward California’s is above average?
(i) Construct a 95% confidence interval for the mean of this sample.
(ii) What is the probability that the average of this sample is between 48 and 52?
d. If California is actually 6 points above average, and you use an alpha of .01 and a sample of 64, what would the study’s power be? If you changed alpha to .05, how would that affect the effect size and the power?
e. Snidely took a sample of 16 Riverdale residents and a sample of 16 University City residents and asked them to rate their cities. The average rating for Riverdale was 60 (standard deviation of 12) and the average rating for University City was 54 (standard deviation of 12). Can you conclude that there is a difference between Riverdale and University City residents?
Explanation / Answer
Q1.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=54
Standard deviation( sd )=12
Sample Size(n)=36
Confidence Interval = [ 54 ± Z a/2 ( 12/ Sqrt ( 36) ) ]
= [ 54 - 1.96 * (2) , 54 + 1.96 * (2) ]
= [ 50.08,57.92 ]
Q2.
Mean ( u ) =50
Standard Deviation ( sd )=12
Normal Distribution = Z= X- u / sd ~ N(0,1)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 48) = (48-50)/12
= -2/12 = -0.1667
= P ( Z <-0.1667) From Standard Normal Table
= 0.43382
P(X < 52) = (52-50)/12
= 2/12 = 0.1667
= P ( Z <0.1667) From Standard Normal Table
= 0.56618
P(48 < X < 52) = 0.56618-0.43382 = 0.1324