Math 283-Autumn 2021 Review Solution (UOWD)
Problem 1
A co¤ee drinker’s organization estimates that the mean consumption of co¤ee by an indi-
vidual in Brazil is more than 8 gallons per year. A random sample of 100 residents in Brazil
is considered and they …nd a mean consumption of co¤ee of 7.9 gallons per year with a
standard deviation of 2.6 gallons.
(a) Find a 93% con…dence interval for the mean consumption of co¤ee.
(b ) At = 0:07; can you support the organization’s claim?
(i) Write the claim mathematically and identify H0 and Ha .
(ii) Find the P-value.
(iii) Decide whether to reject or fail to reject the null hypothesis.
(iv) Interpret the decision in the context of the original claim.
Solution
(a) n = 100 > 30: The con…dence interval is
s s
X z2 p ; X + z2 p
n n
0:07
= 0:07; =
= 0:035; z0:035 = 1:81
2 2
The 93% con…dence interval for the mean consumption of co¤ee is
2:6 2:6
7:9 (1:81) ; 7:9 + (1:81) = [7: 429 4; 8: 370 6]
10 10
(b)
(i)
H0 : 8; Ha : > 8 (claim):
(ii)
X 0 7:9 8
z0 = = = 0:38
ps 2:6
n 10
P value ' 0:64
(iii)
P value ' 0:64 > = 0:07 ) fail to reject H0 :
(iv) At the 7% signi…cance level, there is not enough evidence to support the tea drinker’s
organization’s claim that the mean consumption of co¤ee by an individual in Brazil is more
than 8 gallons per year.
1
Problem 2
A mobile phone repairer claims that the mean repair cost for damaged mobile phones is less
than $100. You are a consumer reporter and you want to test this claim. For this, you
collect a random sample of 5 mobile phones and obtain a mean repair cost of $75 and a
standard deviation of $12.50. At = 0:01; do you have enough evidence to support the
claim? Assuming that the population is normally distributed.
(a) Write the claim mathematically and identify H0 and Ha .
(b) Find the critical value(s) and the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
(d) Interpret the decision in the context of the original claim.
Solution
(a)
H0 : 100; Ha : < 100 (claim):
(b)
t0 = 3:747; Rejection Region: t < 3:747
(c)
tscore = 4:472 < 3:747 ) Reject H0 :
(d) At the 1% signi…cance level, there is su¢ cient evidence to support the the mean repair
cost for damaged mobile phones is less than $100.
2
Problem 3
From long experience, a variable is known to be normally distributed with standard deviation
6 about any given value of the mean, i.e watever the current mean is, the variability about
the mean is constant. A random sample of 16 items from the population has a mean of 53.
At = 0:05; can you say that the current population mean is 50?
(a) Write the claim mathematically and identify H0 and Ha :
(b) Find the critical value(s) and the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
(d) Interpret the decision in the context of the original claim.
Solution
(a)
H0 : = 50 (claim), Ha : 6= 50
(b) Since the population is normally distributed and the standard deviation = 6 is konwn,
we will use a two sided z-test.
0:05
= = 0:025
2 2
z0 = 1:96; z0 = 1:96
Rejection region: z < 1:96 or z > 1:96:
(c)
53 50
zscore = = 2 > 1:96 ) Reject H0 :
p6
16
(d) At the 5% signi…cance level, there is su¢ cient evidence to reject the claim that the
current population mean is 50.
3
Problem 4
The following data represent the ages and the number of hours slept in one night of seven
adults.
Age 35 20 59 42 68 38 75 (x)
Hours slept 7 9 5 6 5 8 4 (y)
(a) Construct a scatter plot for the data. Do the data appear to have a positive linear
correlation, a negative linear correlation, or no linear correlation? Explain.
(b) Calculate the correlation coe¢ cient. What can you conclude?
(c) Test the level of signi…cance of the correlation coe¢ cient. Use = 0:05:
(d) Find the equation of the regression line for the data. Draw the regression line on the
scatter plot.
(e) How many hours of sleep would you predict for an adult of age 45 years?
(f) Find the coe¢ cient of determination and interpret the results.
(g) Construct a 95% prediction interval for the number of hours of sleep for an adult who is
45 years old.
Solution
(a)
Strong Negative Correlation
(b)
r' 0:95 con…rming the strong negative correlation
(c)
H0 : = 0; Ha : 6= 0
= 0:05; df = 7 2 = 5; CV = t5;0:025 = 2:571;
Rejection Region : t < 2:571 or t > 2:571
r 0:95
tscore = q =q = 6: 803 1 < 2:571 ) Reject H0 :
1 r2 1 ( 0:95)2
n 2 5
The correlation is signi…cant.
(d)
y= 0:0865x + 10:4495
4
(e) When x = 45;
y = ( 0:0865) (45) + 10:4495 = 6: 557 (hours)
(f)
r2 = ( 0:95)2 = 0:902 5
) about 90:25% of the variation can be explained and about 9:75% unexplained.
(g)
5
Problem 5
A statistician claims that a population of interest has a mean greater than 6.4. The statis-
tician uses a sample of 16. Assuming that the population is normal, what is the rejection
region for the needed distribution to support the claim. Use = 0:001:
Solution
Because the standard deviation is unkown and the population is normal, we will use the
t-distribution.
Claim: > 6:4:
Complement: 6:4:
H0 : 6:4 Ha : > 6:4:
A right-tailed test is needed.
t0 = tn 1 ; = t15;0:001 = 3:733:
Rejection region is t > 3:733:
6
Problem 6
In an olympiad contest, the mean score for 43 male participants is 21.1 and the standard
deviation is 5. The mean score for 56 female participants is 20.9 and the standard deviation
is 4.7. At = 0:01; can you reject the claim that male and female have equal score?
a. Identify the claim and state H0 and Ha :
b. Find the critical value and identify the rejection region.
c. Find the standardized test statistic.
d. Decide whether to reject the null hypothesis.
e. Interpret the decision in the context of the original claim
Solution
a.
H0 : 1 = 2 (claim), Ha : 1 6= 2
b.
z0 = 2:575 and z0 = 2:575
Rejection Region: Z < 2:575; z > 2:575
c.
T S = 0:202
d. Fail to reject because T S = 0:202 is not in the rejection region.
e. At the 1% signi…cance level, there is insu¢ cient evidence to reject the claim that male
and female participants have equal scores.
7
Problem 7
A manager from Morocco claims that the mean annual income is greater in Casablanca than
it is in Marrakech. In Casablanca, a random sample of 19 residents has a mean annual
income of $42,200 and a standard deviation of $8600. In Marrakech, a random sample of
15 residents has a mean annual income of $37,900 and a standard deviation of $5500. At
= 0:1; can you support the manager’s claim? Assume the population variances are not
equal.
a. Identify the claim and state H0 and Ha :
b. Find the critical value and identify the rejection region.
c. Find the standardized test statistic.
d. Decide whether to reject the null hypothesis.
e. Interpret the decision in the context of the original claim
Solution
a.
H0 : 1 2 , Ha : 1 > 2 (claim)
b. t0 = 1:345; Rejection region t > 1:345:
c. TS = 1:769
d. TS is in the rejection region: reject H0 :
e. At the 10% signi…cance level, there is enough evidence to support the manager’s claim
that the mean annual income is greater in Casablanca than it is in Marrakech. (1pt)
8
Problem 8
The table below shows the average weekly incomes (in Euro) of full time male and female
workers for 5 years. The equation of the regression line is
y^ = 1:369x 402:687
Average weekly incomes of male workers x Average weekly incomes of female workers y
670 512
679 529
695 552
713 573
722 585
(a) Find the coe¢ cient of determination and interpret the result,
(b) Find the standard error of estimate se .
(c) When the average weekly incomes of male workers is $650, …nd a 99% prediction interval
for the median weekly earnings of female workers.
Solution
(a) r2 = 0:994; 99.4% of the variation in the average weekly incomes of female workers can
be explained by the variation in the average weekly incomes of male workers, and 0.6% of
the variation is unexplained.
(b)
Se = 2:719
(c)
463:15 < y < 511:18
You can be 99% con…dent that the average weekly incomes of female workers will be between
$463.15 and $511.18 when the average weekly earnings of male workers is $650.
9
Problem 9
A consumer group, concerned about the mean fat content of a certain grade of steakburger
submits to an independent laboratory a random sample of 12 steakburgers for analysis. The
percentage of fat in each of the steakburgers is as follows.
21 18 19 16 18 24 22 19 24 14 18 15
The manufacturer claims that the mean fat content of this grade of steakburger is less than
20%. Assuming percentage fat content to be normally distributed with a standard deviation
of 3, carry out an appropriate hypothesis test in order to advise the consumer group as to
the validity of the manufacturer’s claim. Use the critical value method with = 0:05:
(a) Write the claim mathematically and identify H0 and Ha :
(b) Find the critical value(s) and identify the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
Solution
(a)
H0 : 0:2 and Ha : < 0:2 (claim)
(b) = 0:05; CV = 1:645; Rejection region is Z < 1:645
(c)
10
Problem 10
An environmentalist estimates that the mean waste recycled by adults in the United States is
more than 1 pound per person per day. You want to test this claim. You …nd that the mean
waste recycled per person per day for a random sample of 12 adults in the United States is
1.46 pounds and the standard deviation is 0.28 pound. At = 0:05 can you support the
claim?
(a) Write the claim mathematically and identify H0 and Ha :
(b) Find the critical value(s) and identify the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
Solution
(a)
(b) Sample size less than 30, we use a t-test.
(c)
X 1:46 1
t= = = 5:691
ps 0:28
p
n 12
5.691 is in the rejection region, we reject the null hypothesis.At the 5% signi…cance level,
there is su¢ cient evidence to support the environmentalist’s claim that the mean waste
recycled by adults in the United States is more than 1 pound per person per day .
11
Problem 11
A marketing manager conducted a study to determine whether there is a linear relationship
between money spent on advertising and company sales. The data are shown in the table
below.
(1) Calculate the correlation coe¢ cient for the advertising expenditures and company sales.
What can you conclude?
(2) Test the signi…cance of this correlation coe¢ cient. Use = 0:05:( State the null and
alternative hypotheses, specify the level of signi…cance, identify the degrees of freedom,
determine the critical values and the rejection regions, make a decision the signi…cance of
this correlation coe¢ cient).
(3) Find the equation of the regression line for the advertising expenditures and company
sales.
(4) Construct a 95% prediction interval for the company sales when! the advertising expenses
X
are $2100. What can you conclude ? (yi y^i )2 = 635:3463
i
Solution
(1)
12
(2)
(3)
13
(4)
14
Problem 12
A survey indicates that the mean per capita credit card charge for residents of New Hamp-
shire and New York is $3900 and $3500 per year, respectively. The survey included a ran-
domly selected sample of size 50 from each state, and sample standard deviations are $900
(NH) and $500 (NY).The two samples are independent. At = 0; 01 is there enough evi-
dence to conclude that there is a di¤erence in the mean credit card charges?
(a) Write the claim mathematically and identify H0 and Ha :
(b) Find the critical value(s) and identify the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
Solution
(a)
(b)
(c) zscore = 2.747 which is in the rejection region. Reject H0 : At the 1% signi…cance level,
there is enough evidence to support the claim that there is a di¤erence in the mean per
capita credit card charges for residents of New Hampshire and New York
15
Problem 13
The braking distances of 8 Volkswagen GTIs and 10 Ford Focuses were tested when traveling
at 60 miles per hour on dry pavement. The results are shown below. Can you conclude that
there is a di¤erence in the mean braking distances of the two types of cars? Use = 0:01:
Assume the populations are normally distributed and the population variances are not
equal.
(a) Write the claim mathematically and identify H0 and Ha :
(b) Find the critical value(s) and identify the rejection region(s).
(c) Decide whether to reject or fail to reject the null hypothesis.
Solution
(a) You want to test whether the mean stopping distances are di¤erent. So, the null and
alternative hypotheses are
(b)
(c)
16
