# Hypothesis Tests One Sample Means

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```					Hypothesis Tests

One Sample Means
2: can I tell if they really
ExampleHowA government agency
are underweight?
complaints that a particular
Hypothesis test
restaurant has been selling
will help me
decide! a sample & find The
underweight hamburgers. x.
Take
patties are “a quarter pound” (4
But how do I know if this x is one
ounces). I expect to happen or is it one
that
that is unlikely to happen?
What are hypothesis tests?
Calculations that tell us if a value
occurs by random chance or not – if
it is statistically significant
Is it . . .
– a random occurrence due to
variation?
– a biased occurrence due to some
other reason?
Nature of hypothesis tests -
How does a murder trial work?
• First begin by supposing the
First assume present
“effect”- is NOTthat the
• Next, see if data provides
person is innocent
Then – must have the
evidence againstsufficient
evidence
suppositionto prove guilty
Hmmmmm …
Example:         murder
Hypothesis tests use   trial
the same process!
Notice the steps are the
hypothesis statements –
which you will learn today
1) Assumptions
2) Hypothesis statements &
define parameters
3) Calculations
4) Conclusion, in context
Assumptions for z-test (t-test):
YEA –
These are the same
•   Have an SRS of context
assumptions as confidence
•                    intervals!!
Distribution is (approximately)
normal
– Given
– Large sample size
– Graph data
• s is known (unknown)
Example 1: Bottles of a popular cola
•Have an SRS to contain
are supposed of bottles 300 mL of
cola. There is some variation from
•Sampling distribution is approximately
bottle to bottle. An inspector, who
normal because the boxplot is
symmetrical
suspects that the bottler is under-
• s is unknown
filling, measures the contents of six
randomly selected bottles. Are the
assumptions met?

299.4 297.7 298.9 300.2 297 301
Writing Hypothesis statements:

• Null hypothesis – is the statement
being tested; this is a statement of
“no effect” or “no difference”
H0:
• Alternative hypothesis – is the
statement that we suspect is true

Ha:
The form:
Null hypothesis
H0: parameter = hypothesized value

Alternative hypothesis
Ha: parameter > hypothesized value
Ha: parameter < hypothesized value
Ha: parameter = hypothesized value
Example 2: A government agency has
particular restaurant has been selling
underweight hamburgers. The
are “a quarter pound” (4 ounces).
State the hypotheses :

H0: m = 4            Where m is the true
mean weight of
Ha: m < 4            hamburger patties
Example 3: A car dealer advertises
that is new subcompact models get
47 mpg. You suspect the mileage
might be overrated.
State the hypotheses :

H0: m = 47
Where m is the
true mean mpg
Ha: m < 47
Example 4: Many older homes have electrical
systems that use fuses rather than circuit
breakers. A manufacturer of 40-A fuses
wants to make sure that the mean amperage at
which its fuses burn out is in fact 40. If the
mean amperage is lower than 40, customers
will complain because the fuses require
replacement too often. If the amperage is
higher than 40, the manufacturer might be
liable for damage to an electrical system due
to fuse malfunction. State the hypotheses :
H0: m = 40             Where m is the true
mean amperage of
Ha: m = 40             the fuses
• ALWAYS refer to populations
(parameters)
• The null hypothesis for the
“difference” between populations is
usually equal to zero
H0: mx-y= 0
• The null hypothesis for the correlation
(rho) of two events is usually equal to
zero.     H0: r= 0
Activity: For each pair of
Must use indicate (population) x
Must be NOT are
hypotheses,parameter whichequal!not
is a explain (sample)
legitimate & statisticswhy
a) H 0 : m  15 ; H a : m  15
 is the population
Ha : x 
b) H 0 : x  123;proportion!123
Must use same
:   . as H0
r is number1; H :   .1
c) H 0parameter fora! population
correlation coefficient – but H0
d) H 0 : m  .4;be “=“ !m  .6
MUST H a :
e) H 0 : r  0 ; H a : r  0
P-values -

• The probability that the test
statistic would have a value as
extreme or more than what
is actually observed
In other words . . . is it far
out in the tails of the
distribution?
Level of Significance
Activity
Level of significance -
• Is the amount of evidence
necessary before we begin to doubt
that the null hypothesis is true
• Is the probability that we will
reject the null hypothesis, assuming
that it is true
• Denoted by a
– Can be any value
– Usual values: 0.1, 0.05, 0.01
– Most common is 0.05
Statistically significant –
• The p-value is as small or smaller
than the level of significance (a)

• If p > a, “fail to reject” the null
hypothesis at the a level.
• If p < a, “reject” the null
hypothesis at the a level.
• ALWAYS make decision about the null
hypothesis!
• Large p-values show support for the
null hypothesis, but never that it is
true!
• Small p-values show support that the
null is not true.
• Double the p-value for two-tail (=)
tests
• Never accept the null hypothesis!
Never “accept” the null hypothesis!

Never “accept” the null
hypothesis!

Never “accept” the null
hypothesis!
At an a level of .05, would you
reject or fail to reject H0 for
the given p-values?
a) .03     Reject
b) .15     Fail to reject
c) .45     Fail to reject
d) .023    Reject
Calculating p-values
• For z-test statistic –
– Use normalcdf(lb,ub)
– [using standard normal curve]

• For t-test statistic –
– Use tcdf(lb, ub, df)
Draw & shade a curve &
calculate the p-value:
1) right-tail test   t = 1.6; n = 20

2) left-tail test    z = -2.4; n = 15

3) two-tail test     t = 2.3; n = 25
Writing Conclusions:
1) A statement of the decision
being made (reject or fail to
AND
2) A statement of the results in
context. (state in terms of Ha)
“Since the p-value < (>) a,
I reject (fail to reject)
the H0. There is (is not)
sufficient evidence to
suggest that Ha.”
Be sure to write Ha in
context (words)!
Example 5: Drinking water is
considered unsafe if the mean
H0: m = 15
concentration of lead is 15 ppb (parts
Ha: billion)
per m > 15 or greater. Suppose a
t=2.1
Where m is the true mean concentration
community randomly selects 25 water
of lead in drinking water a t-test
samples and computesreject H0. There is
Since the p-value < a, I
P-value of 2.1. Assume that
statistic = tcdf(2.1,10^99,24) that the
mean concentration normally
concentrations are of lead in drinking
=.0232
water is greater than 15 hypotheses,
distributed. Write the ppb.
calculate the p-value & write the
appropriate conclusion for a = 0.05.
Example 6: A certain type of frozen
dinners states that the dinner
contains 240 calories. A random
H0: m = 240
sample: of> 12 of these frozen dinners
Ha m 240
Where m is thet=1.9 mean caloric see
was selected fromtrue  production to
Since the p-value < a, I reject H0. There is
content of the frozen dinners
if the caloric content was greater
sufficient evidence to suggest that the
than stated on the box. The t-test
P-value = caloric content of
true meantcdf(1.9,10^99,11) these frozen
=.0420 was calculated to be 1.9.
statistic is greater than 240 calories.
dinners
Assume calories vary normally. Write
the hypotheses, calculate the p-value
& write the appropriate conclusion
for a = 0.05.
Formulas:
s known:
statistic - parameter
test statistic 
standard deviation of statistic

z=              x m
σ
n
Formulas:
s unknown:
statistic - parameter
test statistic 
standard deviation of statistic

t=              x m
s
n
Example 7: The Fritzi Cheese Company buys milk from
several suppliers as the essential raw material for its
cheese. Fritzi suspects that some producers are
adding water to their milk to increase their profits.
Excess water can be detected by determining the
freezing point of milk. The freezing temperature of
natural milk varies normally, with a mean of -0.545
degrees and a standard deviation of 0.008. Added
water raises the freezing temperature toward 0
degrees, the freezing point of water (in Celsius). The
laboratory manager measures the freezing
temperature of five randomly selected lots of milk
from one producer with a mean of -0.538 degrees. Is
there sufficient evidence to suggest that this
producer is adding water to his milk?
SRS?
Assumptions:
Normal?
•I have an SRS of milk from one producer             How do you
•The freezing temperature of milk is a normal distribution.
know?
(given)
• s is known                  Do you
H0: m = -0.545              know s?              What are your
hypothesis
Ha: m > -0.545                                  statements? Is
where m is the true mean freezing temperature of milk word?
there a key

 .538   .545
z                    1.9566                   Plug values
.008                                    into formula.
5
p-value = normalcdf(1.9566,1E99)=.0252
Use normalcdf to
a = .05                                     calculate p-value.
a & make decision
Since p-value < a, I reject the null hypothesis.
There is sufficient evidence to suggest that the true mean freezing
temperature is greater than -0.545. This suggests that the
producer is adding water to the milk.

Write conclusion in
context in terms of Ha.
Example 8: The Degree of Reading Power
(DRP) is a test of the reading ability of
children. Here are DRP scores for a random
sample of 44 third-grade students in a
suburban district:
(data on note page)
At the a = .1, is there sufficient evidence to
suggest that this district’s third graders
reading ability is different than the national
mean of 34?
SRS?
• I have an SRS of third-graders
Normal?
•Since the sample size is large, the sampling distribution is
approximately normally distributed                       How do you
know?
OR
Do you
know s?
•Since the histogram is unimodal with no outliers, the are your
What
sampling distribution is approximately normally distributed
hypothesis
• s is unknown                                       statements? Is
H0: m = 34     where m is the true mean reading there a key word?
Ha: m = 34     ability of the district’s third-graders
35.091  34                                     Plug values
t              .6467
11.189                                         into formula.
44
p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212
Use tcdf to calculate
a = .1                                         p-value.
a & make decision
Since p-value > a, I fail to reject the null hypothesis.
There is not sufficient evidence to suggest that the true mean
national mean of 34.

Write conclusion in
context in terms of Ha.
Example 9: The Wall Street Journal
(January 27, 1994) reported that based
on sales in a chain of Midwestern grocery
stores, President’s Choice Chocolate Chip
Cookies were selling at a mean rate of
\$1323 per week. Suppose a random sample
of 30 weeks in 1995 in the same stores
showed that the cookies were selling at
the average rate of \$1208 with standard
deviation of \$275. Does this indicate that
the sales of the cookies is different from
the earlier figure?
Assume:
•Have an SRS of weeks
•Distribution of sales is approximately normal due to large
sample size
• s unknown
H0: m = 1323     where m is the true mean cookie sales per
Ha: m ≠ 1323          week
1208  1323
t              2.29    p  value  .0295
275
30
Since p-value < a of 0.05, I reject the null hypothesis. There
is sufficient evidence to suggest that the sales of cookies are
different from the earlier figure.
Example 9: President’s Choice Chocolate
Chip Cookies were selling at a mean rate of
\$1323 per week. Suppose a random sample
of 30 weeks in 1995 in the same stores
showed that the cookies were selling at the
average rate of \$1208 with standard deviation
of \$275. Compute a 95% confidence interval
for the mean weekly sales rate.
CI = (\$1105.30, \$1310.70)
Based on this interval, is the mean weekly
sales rate statistically different from the
reported \$1323?
What do you notice about the decision from the
confidence interval & the hypothesis test?
What decision would you make on Example 9 if a =
.01?

What confidence level would be correct to use?

Does that confidence interval provide the same
decision?

If Ha: m < 1323, what decision would the
hypothesis test give at a = .05? a = .01?

Now, what confidence levels are appropriate for
this alternative hypothesis?
Matched Pairs Test
A special type of t-
inference
Matched Pairs – two forms
• Pair individuals by       • Individual persons
certain characteristics     or items receive both
• Randomly select             treatments
treatment for             • Order of treatments
individual A                are randomly
• Individual B is             assigned or before &
assigned to other           after measurements
treatment                   are taken
• Assignment of B is        • The two measures
dependent on                are dependent on
assignment of A             the individual
Is this an example of matched pairs?

1)A college wants to see if there’s a difference in
time it took last year’s class to find a job after
graduation and the time it took the class from five
years ago to find work after graduation.
Researchers take a random sample from both
classes and measure the number of days between
graduation and first day of employment

No, there is no pairing of individuals, you have
two independent samples
Is this an example of matched pairs?
2) In a taste test, a researcher asks people in a
random sample to taste a certain brand of spring
water and rate it. Another random sample of
people is asked to taste a different brand of water
and rate it. The researcher wants to compare these
samples

No, there is no pairing of individuals, you have
two independent samples – If you would have
the same people taste both brands in random
order, then it would be an example of matched
pairs.
Is this an example of matched pairs?
3) A pharmaceutical company wants to test its new
weight-loss drug. Before giving the drug to a
random sample, company researchers take a
weight measurement on each person. After a
month of using the drug, each person’s weight is
measured again.

Yes, you have two measurements that are
dependent on each individual.
A whale-watching company noticed that many
customers wanted to know whether it was better to
book an excursion in the morning or the afternoon.
To test this question, the company collected the
You may subtract either
way – data careful randomly selected days over
followingjust be on 15 when
writing Ha
the past month. (Note: days were not consecutive.)

Day        1 2    3   4   5   6   7     8     9   10   11 12 13 14 15
Morning    8 9    7 9     10 13 10      8     2   5    7 7 6 8 7
After-
noon
8 10   9 8     9   11 8      10    4   7    8 9 6 6 9

Since you have two values for         First, you must find the
each day, they are dependent on          differences for each
the day – making this data                    day.
matched pairs
Day         1 2    3   4    5   6    7    8    9   10   11 12 13 14 15
Morning     8 9    7   9 10 13 10         8    2   5    7 7 6 8 7
After-
noon
8 10   9   8 9      11 8    10 4 7 8 9 6 6 9
I subtracted:
Differenc
es
0 -1   -2 1 1       2 2     –     -2 -2
Morning -2afternoon -1 -2 0 2 -2

Assumptions:             You could subtract the other way!

• Have an SRS of days for whale-watching
You need to state assumptions using the
• s unknown           differences!
•Since the normal probability plot is approximately linear, the
distribution of difference is approximately normal.
Notice the granularity in this
plot, it is still displays a nice
linear relationship!
Differenc
es
0 -1   -2 1 1    2   2    -2   -2 -2    -1 -2 0 2 -2

Is there sufficient evidence that more whales are sighted in the
afternoon?
H0: mD = 0              If you subtract afternoonhow you
subtracted: M-A
then Ha: m >0
Notice we used mD for differences the
If afternoon isDmore should
Ha: mD < 0                            differences should -?
& it equals 0 since the null be + or be
that there is NO difference.
Don’t look at numbers!!!!
Where mD is the true mean difference
in whale sightings from morning
minus afternoon
Differenc
es
0 -1   -2 1 1    2   2     -2   -2 -2    -1 -2 0 2 -2

finishing the hypothesis test:
xμ         .4  0                         perform a t-test
t                      .945              Notice that if you
using the differences
s      1.639                            subtracted A-M,
n            15                             (L3)
p  .1803                                         statistic
t = + .945, but p-
df  14           α  .05                   value would be the
same
Since p-value > a, I fail to reject H0. There is insufficient
evidence to suggest that more whales are sighted in the
afternoon than in the morning.

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