Download Review Chapter 5 and 6

Review Chapter 9 and 10 Key
1. Find z
2. Find t
2
2
for the 96% confidence interval.
z 2  2.05
for the 95% confidence interval with
n  21 .
t 2  2.086
3. A poll of 1027 adults in a certain country found that 23% identified themselves as the
followers of some religion. The margin of error was 4 percentage points with 99% confidence.
Which of the following represents a reasonable interpretation of the survey results?  (d)
(a) In 99% of samples of adults in a certain country, the proportion who identify themselves
as the followers of some religion is between 19% and 27%
(b) There is between 95% and 100% confidence that 23% of adults in a certain country
identify themselves as the followers of some religion.
(c) There is 99% confidence that 23% of adults in a certain country identify themselves as
the followers of some religion.
(d) There is 99% confidence that the proportion of adults in a certain country who identify
themselves as the followers of some religion is between 19% and 27%.
4. A medical researcher wishes to determine the percentage of females who take vitamins. He
wishes to be 99% confident that the estimate is within 2 percentage points of the true
proportion. A recent study of 180 females showed that 25% took vitamins.
a) How large should the sample size be?
pˆ  0.25; 1  pˆ  0.75
z 
ˆ (1  pˆ )   /2 
Formula to find the sample size: n  p
 E 
2
2
 2.575 
n  (0.25)(0.75) 
  3108.1  3109
 0.02 
b) If no estimate of the sample proportion is available, how large should the sample be?
2
z 
n  pˆ (1  pˆ )   /2 
 E 
ˆ (1  pˆ )  0.25
Since the p̂ is not available, use p
2
 2.575 
n  0.25 
  4144.1  4145
 0.02 
5. Determine the point estimate of the population mean and margin of error for the confidence
interval.
Lower bound is 19, upper bound is 27
a) The point estimate of the population mean is 23 .
b) The margin of error for the confidence interval is 4.
19  27
 23
2
23 – 19 = 4 Or 27 – 23 = 4
6. The data represent a sample of the number of home fires started by candles for the past
several years. Find the 99% confidence interval for the mean number of home fires started
by candles each year. Assume that the data come from a population that is normally
distributed and the sample size does not contain any outliers.
5460
5900
6090
6310
7160
8440
9930
Confidence level  0.99, x  7041.4 s  1610.3 n  7 d . f  7  1  6
From table VI , t  3.707
2
 s 
 s 
x  t 
    x  t 

n
n
2 
2 
 1610.3 
 1610.3 
7041.4  3.707 
    7041.4  3.707 

7 
7 


4785.2    9297.6
The 99% confidence interval for the mean population is 4785.2    9297.6 .
7. A simple random sample of n = 61 is drawn from a population. The sample mean is found to
be x = 125.5 and the sample standard deviation is found to be s = 10.5. Construct a 98%
confidence interval for the population mean. (Round to two decimal places as needed.)
No information is given about the value of the population standard deviation, so
unknown. Because the sample size is big, compute a t-interval.
 is
Confidence level  0.98, x  125.5, s  10.5 n  61 d . f  61  1  6
From table VI , t  2.390
2
 s 
 s 
x  t 
    x  t 

n
n
2 
2 
 10.5 
 10.5 
125.5  2.390 
    125.5  2.390 

 61 
 61 
122.29    128.71
8. A CBS News/ New York Times poll found that 329 out of 763 adults said they would travel to
outer space in their lifetime, given the chance. Estimate the true proportion of adults who
would like to travel to outer space with 92% confidence. (Round to four decimal places as
needed.)
x  329, n  763
x 329
pˆ  
 0.43
n 763
1  pˆ  0.57
Confidence Interval for a proportion:
pˆ  z /2
pˆ (1  pˆ )
pˆ (1  pˆ )
 p  pˆ  z /2
n
n
z /2  1.75 for 92%
0.43(0.57)
0.43(0.57)
 p  0.43  1.75
763
763
0.3986  p  0.4613
0.43  1.75
9. A researcher wishes to estimate, within $300, the average amount of money a county spends
on road repairs each year. If she wants to be 90% confident, how large a sample is needed?
The standard deviation of all road repair spending is known to be $900, and the population is
normally distributed.
E  300, za  1.645 (  0.10),   900
2
 za  
n 2
 E

2
  1.645  900 2
 
 24.3542  n  25
  300 

10. It is desired to estimate the mean number of chocolate chips per cookie for a large national
brand. How many cookies would have to be sampled to estimate the true mean number of
chips per cookie within 2 chips with 98% confidence? Use s  10.1 chips.

0.02
 0.01
2
2
1  0.01  0.99  ( Z / 2 )  2.33
z  2.33, s  10.1, E  2
  0.02 

2
2
 z s   2.33(10.1) 2
n 2  
  138.45  139
 E  
2



Therefore, to be 98% confident that the estimate is within 2 chips of the true mean number of
chocolate chips per cookies, at least 139 cookies have to be sampled.
11. Fill in the blanks to complete the statement.
(a) The null hypothesis is a statement of no change, no effect, or no difference.
(b) If we reject the null hypothesis when the statement in the null hypothesis is true, we have
made a Type I error.
12. According to the report, the mean monthly cell phone bill was $48.29 three years ago. A
researcher suspects that the mean monthly cell phone bill is different today.
(a) State the null and alternative hypotheses.
H 0 :   $48.29
H1 :   $48.29
(b) Explain what it would mean to make a Type I error. Choose the correct answer below.
a
a. The sample evidence led the researcher to believe the mean monthly cell phone bill
is different from $48.29, when in fact the mean bill is $48.29.
b. The sample evidence led the researcher to believe the mean monthly cell phone bill
is higher than $48.29, when in fact the man bill is $48.29.
c. The sample evidence did not lead the researcher to believe the mean monthly cell
phone bill is different from $48.29, when in fact the mean bill is different from $48.29.
(c) Explain what it would mean to make a Type II error. Choose the correct answer below.
a
a. The sample evidence did not lead the researcher to believe the mean monthly cell
phone bill is different from $48.29, when in fact the mean bill is different from $48.29.
b. The sample evidence did not lead the researcher to believe the mean monthly cell
phone bill is higher than $48.29, when in fact the mean bill is higher than $48.29.
c.
The sample evidence led the researcher to believe the mean monthly cell phone bill
id different from $48.29, when in fact the mean bill is different from $48.29.
13. Suppose the null hypothesis is not rejected. State the conclusion based on the results of the
test.
Three years ago, the mean price of a single-family home was $243,723. A real estate
broker believes that the mean price has decreased since then.
Which of the following is the correct conclusion?  c
a. There is not sufficient evidence to conclude that the mean price of a single-family
home has not changed.
b. There is sufficient evidence to conclude that the mean price of a single-family home
has decreased.
c. There is not sufficient evidence to conclude that the mean price of a single-family
home has decreased.
d. There is sufficient evidence to conclude that the mean price of a single-family home
has not changed.
14. Find the p-value for a two tailed test with observed
15. Find the p-value for a left tailed test with observed
z  2.06 . p-value = 0.0394
t  2.602 and n  16 . p-value = 0.01
16. In a previous year, 48% of females aged 15 years of age and older lived alone. A sociologist
tests whether this percentage is different today by conducting a random sample of 1000
females aged 15 years of age and older and finds that 440 are living alone. Is there sufficient
evident at the   0.05 level of significance to conclude the proportion has changed?
Assume that the sampling distribution of p̂ is approximately normal.
p=0.48, n=1000, x=440,   0.05 , pˆ 
x 440

 .440
n 1000
a. State the null and alternative hypotheses for the test.
H 0 : p  0.48
H1 : p  0.48 (a two-tailed test)
b. Compute the test value.
z0 
c.
pˆ  p0

p0 (1  p0 )
n
0.440  0.48
 2.53
(0.48)(1  0.48)
1000
Compute the p-value.
Find the area that corresponds to z = - 2.53. The area is 0.0057.
So, the P-value is 2(0.0057)=0.0114
d. Make a decision whether to reject or not to reject the null hypothesis.
Is p-value   ? Yes, we reject the null hypothesis.
e. Summarize the conclusion.
There is enough evidence at
changed.
  0.05 to support the claim that the proportion has
17. In 1997, 4% of mothers smoked more than 21 cigarettes during their pregnancy. An
obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less
than 4% today. She randomly selects 120 pregnant mothers and finds that 3 of them smoked
21 or more cigarettes during pregnancy. Does the sample data support the obstetrician’s
belief? Use the classical approach with   0.01 level of significant. Assume that the
sampling distribution of p̂ is approximately normal.
p=0.04, n=120, x=3,
  0.01 , pˆ 
x
3

 0.025
n 120
a. State the null and alternative hypotheses for the test.
H 0 : p  0.04
H1 : p  0.04 (a left-tailed test)
b. Compute the test value.
z0 
c.
pˆ  p0
0.025  0.04

 0.84
p0 (1  p0 )
(0.04)(0.96)
120
n
Compute the critical value.
Use
z  2.33
d. Make a decision whether to reject or not to reject the null hypothesis.
Is the test value less than the critical value? In other words, does the test statistic fall in
the critical region? No, do not reject H 0 .
e. Summarize the conclusion.
There is not enough evidence to support the claim that the percentages of mothers who
smoke 21 cigarettes or more is less than 4%.
18. The mean height of American males is 69.5 inches. The heights of the 43 male U.S.
presidents (Washington through Obama) have mean 70.78 inches and a standard deviation
of 2.77 inches. Treating the 43 presidents as a simple random sample, determine if there is
evident to suggest that U.S presidents are taller than the average American male. Use the
  0.05 level of significance. Use the p-value method. Assume that the sample is normally
distributed.
a. State the null and alternative hypotheses for the test.
H 0 :   69.5
H1 :   69.5 (a right-tailed test)
b. Compute the test value.
t0 
c.
x  0 70.78  69.5

 3.030
s
2.77
n
43
Compute the p-value.
d.f = 43 –1 = 42  Use d.f = 40
0.001  P-value  0.0025
d. Make a decision whether to reject or not to reject the null hypothesis.
Is p-value   ? Yes, we reject the null hypothesis.
e. Summarize the conclusion.
There is enough evidence at the   0.05 level of significance to conclude that U.S.
presidents are taller than the typical American male.
19. A physician claims that joggers’ maximal volume oxygen uptake is greater than the average
of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per kilogram (ml/kg) and a
standard deviation of 6 ml/kg. If the average of all adults is 36.7 ml/kg, is there enough
evidence to support the physician’s claim at   0.05? Use the p-value method.
Assume that the sample is normally distributed.
a. Identify the claim and state the null and alternative hypotheses for the test.
H 0 :   36.7
H1 :   36.7 (a right-tailed test)
b. Compute the test value.
n  15 x  40.6 s  6   0.05
x  0 40.6  36.7
test value : t0 

 2.52
s
6
n
15
c.
Find the P-value.
0.01  p-value  0.02
d.f = n –1 = 14
d. Make a decision whether to reject or not to reject the null hypothesis.
Is p-value   ?
Yes, we reject the null hypothesis.
e. Summarize the conclusion.
There is enough evidence to support the claim that joggers’ maximal volume oxygen
uptake is greater than the average of all adults.
20. Explain what a P-value is. Choose the correct answer below.
b
a. A P-value is the number of standard deviations that the observed proportion is from the
proportion stated in the null hypothesis.
b. A P-value is the probability of observing a sample statistics as extreme or more extreme
than the one observed under the assumption that the statement in the null hypothesis is
true.
c. A P-value is the value used to designate the area  in either the left- or right-tail of the
normal curve.