A news paper article says that on the average, collage freshman spend 7.5 hours a week going to parties. One administrator does not believe that these figures apply at her college, which has nearly 3,000 freshman. She takes a simple random sample of 100 freshmen, and interviews them. On average, they report 6.6 hours a week going to parties, and the SD is 9 hours. Is the difference between 6.6 and 7.5 real? a) Formulate the null and alternative hypotheses in terms of a box model. b) Fill in the bla

. The null says that the average of the box is ?. The alternative says that average of the box is ?. c) Now answer the question: is the difference real?

. The null says that the average of the box is ?. The alternative says that average of the box is ?. c) Now answer the question: is the difference real?

2 Answers

Best Answer

Step 1
Given: Average sample =6.6
SD sample =9
Number of draws =100
a) We will make 100 draws from a box containing a ticket per freshman, while the ticket contains the number of hours spend a week going to parties.
The null hypothesis states that the average of the box is equal the claimed average of 7.5 hours, while alternative hypothesis states that the average of the box is less than 7.5 hours.
Step 2
b) We will make 100 draws from a box containing a ticket per freshman, while the ticket contains the number of hours spend a week going to parties.
The null hypothesis states that the average of the box is equal the claimed average of 7.5 hours, while the alternative hypothesis states that the average of the box is less than 7.5 hours.
Step 3
c) The standard error of the sum is the product of the square root of the number of draws and the standard deviation of the sample.
SE sum=sqrt(text(Number of draws)) * SD sample
=sqrt(100) * 9
=10 * 9
=90
The standard error of the average is the standard error of the sum divided by the number of draws:
SE average=(SE sum)/(text(Number of draws))
=(90)/(100)
=0.90
Step 4
Chance
mu=Mead=7.5
sigma=Standard deviation=0.90
x=6.6
The z-score is the value decreased by the mean, divided by the standard deviation.
z=(x-mu)/(sigma)=(6.6-7.5)/(0.90)=-1.00
The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z, assuming that the null hypothesis is true. P(-1.00<Z<1.00) is given in the row starting with 1.00.
P=P(Z<-1.00)=(1)/(2)P(Z<-1.00 or Z>1.00)
=(1)/(2)(100%-P(-1.00<Z<1.00))
=(1)/(2)(100%-68.27%)
=(1)/(2)(31.73%)
=15.865%
Since the P-value is not small, it is not unusual to obtain a sample average of 6.6 when the actual average is 7.5 and thus the difference does not appear real (as it appears to be due to chance variation).

Best Answer

a. The hypothesis Null and alternative can be formulate as a:

H0=mu=7.5 hr/week

H_(a)=mu<7.5 hr/week

b. The Null hypothesis is true and the administrator believes that the college freshman dot not spend 7.5 hours

E(x) = mu = 7.5 hr/week

SE (x) = (sigma)/(sqrt(n)) = (9)/(sqrt(100))

SE (x) = 0.9

Z' = 6.6 - 7.5 / SE

Z' = 6.6 -7.5 / 0.9 = -1

The empirical rule in this case is use

% = (100-68)/(2) = 16%

c. There is not strong arguments of Null Hypothesis so can be accept the H0 and is no significantly different from the report of 7.5 hr / week

H0=mu=7.5 hr/week

H_(a)=mu<7.5 hr/week

b. The Null hypothesis is true and the administrator believes that the college freshman dot not spend 7.5 hours

E(x) = mu = 7.5 hr/week

SE (x) = (sigma)/(sqrt(n)) = (9)/(sqrt(100))

SE (x) = 0.9

Z' = 6.6 - 7.5 / SE

Z' = 6.6 -7.5 / 0.9 = -1

The empirical rule in this case is use

% = (100-68)/(2) = 16%

c. There is not strong arguments of Null Hypothesis so can be accept the H0 and is no significantly different from the report of 7.5 hr / week