Download MA 120 Quiz Three

1. Cheese consumption in the United States was 28.0 pound per person in 1997. Figures
for last year’s cheese consumption are given below. At the .10 confidence level, has per
capita cheese consumption in the U. S. increased since 1997. (Assume that σ = 6.9
pounds.)
40
27
30
38
31
23
26
26
27
31
27
30
39
20
30
32
22
18
31
16
36
41
33
21
38
34
20
22
25
30
28
36
27
30
23
Please refer to EXCEL file for calculation (sheet2)
The sample mean value is X=28.8
To test (At the .10 significance level- There is obviously a typo in the question, although
it shows “.10 confidence level”, the confidence level can never be that low. Instead, it
should be the significance level):
Ho: M=28.0
Ha: M>28.0
Standard Error: SE= σ/SQRT(N) = 6.9/SQRT(35)=1.166
Then calculate z=(X-M)/SE = (28.8-28)/1.166= 0.686
From a z table, the 90% critical z value is 1.28
Since the calculated z is smaller than the critical z value, we fail to reject Ho, and
conclude that per capita cheese consumption in the U. S. has NOT increased since 1997
2. The FBI reports that purse snatching in 1998 resulted in a mean loss per victim of
$362. A random sample of twelve victims of purse snatching victims in the current year
is surveyed. The sample mean loss is $314.10 and the sample standard deviation is
$86.90. At the 0.05 level, can we assert that the mean loss for purse snatching victims
has decreased since 1998?
Ho: M=362
Ha: M<362
Standard Error: SE= σ/SQRT(N) = 86.90/SQRT(12)=25.09
Then calculate t=(X-M)/SE = (314.1-362)/ 25.09 = -1.909, and |t|=1.909
From a t table, the 95% one tailed critical t value with d.f=11 is t*=1.796
Since the absolute value of calculated t is greater than the critical t value, we have to
reject Ho, and conclude that the mean loss for purse snatching victims has decreased
since 1998.
3. According to Nielsen Media Research, in 1998, during the time slot from 8:00 PM to
11:00 PM, the average person watched 7 hours and 43 minutes of TV per week. A
random sample of 40 women, age 18 – 24, reveals that their mean viewing time during
the time slot from 8:00 PM to 11:00 PM was 343.25 minutes with a sample standard
deviation of 222.2 minutes. At the 0.05 level, does the sample indicate that women in the
18 – 24 age group spend less time than the average person watching TV in this time slot?
average person watched TV for 7 hours and 43 minutes = 7*60+43=463 minutes
Ho: M=463
Ha: M<463
Standard Error: SE= σ/SQRT(N) = 222.2/SQRT(40)=35.19
Then calculate t=(X-M)/SE = (343.25-463)/ 35.19 = -3.40, and |t|=3.40
From a t table, the 95% one tailed critical t value with d.f=39 is t*=1.684
Since the absolute value of calculated t is greater than the critical t value, we have to
reject Ho, and conclude that women in the 18 – 24 age group spend less time than the
average person watching TV in this time slot.
4. The table that follows gives the distribution of the maximum wind speed of 1,293
tornadoes. Using a 0.01 level of significance, test the hypothesis that 50% of the
population of tornadoes has wind speeds less than 72 mph.
Wind Speed
Number
< 72
73 to 112
113 to 157
158 to 206
207 to 260
260 to 319
696
411
129
43
13
1
Rewrite the table into:
Wind Speed
< 72
>=72
Number
696
597
H0: The data follow a specified distribution.
H1: The data do not follow the specified distribution.
In general, the chi-square test statistic has the form:
X2=Σ(Observed-Expected) 2/Expected =0.5862
And degree of freedom is df=2-1=1
From the X2 table, we can find the 0.01 level of significance X21=6.6349
Since the calculated X2 < X21, we fail to reject H0 at 0.01 level of significance. So we
conclude that 50% of the population of tornadoes has wind speeds less than 72 mph.
5. A computer systems engineer has researched data for e-mail messages handled by a
particular server in a single weekday. The data from a sample of 22 randomly selected
weekdays indicates that the sample mean number of e-mails per day was 41354.136.
The sample standard deviation was 7912.235 e-mails per day. At the 0.05 level of
confidence, does that data indicate that the population standard deviation is no greater
than 5000 e-mails per weekday?
A chi-square test can be used to test if the standard deviation of a population is equal to a
specified value. This is a one-sided test which only tests if SD<=5000. The chi-square
hypothesis test is defined as:
H0: SD=5000
H0: SD<5000
Test Statistic: T =
= (22-1)*(7912.235/5000)^2=52.59
where N is the sample size and is the sample standard deviation. The key element of
this formula is the ratio
, which compares the ratio of the sample standard deviation
to the target standard deviation. The more this ratio deviates from 1, the more likely we
are to reject the null hypothesis.
From the X2 table, we can find the 0.95 level of significance(because this is a lower tail
test) X221=11.59
Since the calculated X2 > X221, we fail to reject H0 at 0.01 level of significance(again,
this is a lower tail test). So we conclude that the population standard deviation is greater
than 5000 e-mails per weekday