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Transcript
```Assignment 3 week 9-week 11
Student Full Name:___________________________________ .
Student ID:__________________________________________ .
CRN No:____________________________________________ .
Branch: _____________________________________________.
STATISTICS
(STAT-101)
Total Points
True/False
____/6
MCQ
____/6
____/18
Total
____/30
Good Luck
STATISTICS (STAT-101)
Marks- 30
Answer all the Questions on the same question paper.
Section-I
State whether the following statements are True or False. (6 marks, 1 Mark
Each)
1. A type I error is the mistake of rejecting the null hypothesis when it is
actually true. True
2. If P βvalue is greater than Ξ± (level of significance) reject the null
hypothesis. False
3. Two samples are dependent if the sample values are matched pairs. True
4. In case of hypothesis testing for a sample, the t statistic is used if π is not
known and sample size n is greater than 30 (n > 30). True
5. A claim that two population proportions are equal, each of the two samples
must satisfy the requirement that ππ β₯ 5 and ππ β₯ 5. True
6. In an unpaired samples t-test with sample sizes π1 = 21 and π2 = 11, the
value of t should be obtained at 10 degree of freedom. True
Section-II
(Multiple Choice Questions)
(6 marks, 1 Mark Each)
1. Area of the rejection region depends on β¦
a. Size of Ξ±
b. Size of Ξ²
c. Test-Statistic.
d. Number of values
2. A two-tailed test is one where:
a. results in only one direction can lead to rejection of the null hypothesis
b. negative sample means lead to rejection of the null hypothesis
c. results in either of two directions can lead to rejection of the null
hypothesis
d. no results lead to the rejection of the null hypothesis
Page 2 of 10
3. A decision in a hypothesis test can be made by using a
a. P-value
b. Critical Value
c. A and B
d. None of the above
4. When carrying out a large sample test of H0: π0 = 10 vs. Ha: π0 > 10 by
using a rejection point, we reject H0 at level of significance πΌ when the
calculated test statistic is:
a. Less than π§πΌ
b. Less than - π§πΌ
c. Greater than π§πΌ/2
d. Greater than π§πΌ
5. A randomly selected sample of 500 college students was asked whether
they had ever used the drug Ecstasy. Sixteen percent (16% or 0.16) of the
500 students surveyed said they had. Which one of the following
statements about the number 0.16 is correct?
a. It is a population proportion.
b. It is a margin of error.
c. It is a sample proportion.
d. It is a randomly chosen number.
6. Two teams of workers assemble automobile engines at a manufacturing
plant in Michigan. A random sample of 145 assemblies from team 1
shows 130 acceptable assemblies. A similar random sample of 125
assemblies from team 2 shows 120 acceptable assemblies. The pooled
proportion for acceptable assemblies is:
250
a.
b.
c.
d.
270
270
250
20
270
270
20
Part-II (Multiple Choice Questions)
MCQ
1
2
3
4
C
C
D
(6 marks, 1 Mark Each)
5
C
Page 3 of 10
6
A
Section βIII
Answer the following Essay Type Questions
(18 marks, 3 Mark Each)
1. A medical researcher claims that less than 20% of American adults are
allergic to a medication. In a random sample of 100 adults, 15% say they
have such an allergy. Test the researcherβs claim at ο‘ = 0.01.
Solution: The products ππ = 100 × 0.20 = 20 and ππ = 100 ×
0.80 = 80 are both greater than 5. So, we can use the z-test. The claim
is βless than 20% are allergic to a medication.β
Step 1
The claim is π < 0.2
Step 2
Alternative to claim is
π β₯ 0.2
Step 3
the null and alternative hypothesis are:
π»0 : π = 0.2.
π»π : π < 0.2 (Claim)
Step 4
the level of significance is ο‘ = 0.01
Step 5
Using the z-test, the standardized test statistic is:
zο½
pΛ ο­ p
ο½
pq
n
0.15 ο­ 0.20
ο» ο­1.25
(0.20)( 0.80)
100
Step 6:
Because the test is a left-tailed test and level of significance is ο‘ = 0.01,
the critical value is π0 = β2.33 and the rejection region is the area where
π < β2.33.
Page 4 of 10
The graph shows the location of the rejection region and the
standardized test statistic. Because z is not in the rejection region, you
should decide fail to reject the null hypothesis. Or P-value is the area to
the left of z = -1.25; The P-value of 0.1056 is greater than the significance
level of ο‘ = 0.01, fail to reject the null Hypothesis.
In other words, there is not enough evidence to support the claim that less
than 20% of Americans are allergic to the medication.
2. Suppose we would like to determine if the typical amount spent per
customer for dinner at a new restaurant in town is more than \$20.00. A
sample of 49 customers over a three-week period was randomly selected
and the average amount spent was \$22.60. Assume that the standard
deviation ο³ is known to be \$5.50. Using a 0.05 level of significance,
would we conclude the typical amount spent per customer is more than
\$20.00?
Solution: ο³ is known ( \$5.50), sample size is 49 (n > 30)
Step 1
The claim is π > 20
Step 2
Alternative to claim is
π β€ 20
Step 3
The null and alternative hypothesis are:
π»π: π = 20, π»π: π > 20
Step 4
the significance level is ο‘ = 0.05
Page 5 of 10
Step 5
ο³ = 5.5, π = 49
π₯Μ = 22.60, π = 20
Zο½
xο­ο­
ο³
n
ο½
22.60 ο­ 20
ο½ 3.30
5.50
49
Step 6:
Right-tailed test, so P-value is the area is to the right of z = 3.30;
The P-value of 0.0005 is less than the significance level of ο‘ = 0.05,
reject the null Hypothesis.
P-value = 0.0005
ο­ = 20
or
z=0
z = 3.30
There is sufficient evidence to conclude the typical amount spent per
customer is more than \$20.00.
3. The scores on an aptitude test required for entry into a certain job position
have a mean at most 500. If a random sample of 36 applicants have a
mean of 546 and a standard deviation of 120, is there evidence that their
mean score is different from the mean that is expected from all
applicants?. Use a 0.05 level of significance.
Solution: population standard deviation is not known, sample size is 36
(n > 30)
Page 6 of 10
Step 1
The claim is π β€ 500
Step 2
Alternative to claim is
π > 500
Step 3
Null and Alternative Hypothesis are:
π»0 : π = 500
π»π : π > 500
.
Step 4
the significance level is ο‘ = 0.05
Step 5
π  = 120, π = 36
π₯Μ = 546, π = 500
tο½
x ο­ ο­ 546 ο­ 500
ο½
ο½ 2.3
s
120
n
36
Step 6
The degrees of freedom (df=n-1) is 36-1=35. Using t-table with 35 degrees
of freedom and level of significance 0.05, the t-value is 1.690. t = 2.3 fall
in the critical region bounded by t = 1.690. Reject the null hypothesis.
ο­ = 500
or
z=0
t = 2.3
Critical value t = 1.690
Page 7 of 10
There is sufficient evidence to conclude that the mean score is different from
the mean that is expected from all applicants.
4. Suppose a researcher believes that college faculty vote at a lower rate than
college students. She collects data from 200 college faculty and 200 college
students using simple random sampling. If 120 of the faculty and 150 of
the students voted in the 2016 Presidential election, is there enough
evidence at the 5% level of significance to support the researcherβs claim?
Solution: two samples are independent and
π₯1 = 120, π1 = 200.
π₯2 = 150, π2 = 200.
Consider
p1: proportion of successes for population 1(Faculty)
p2: proportion of successes for population 2(Students)
Step 1
The claim is π1 < π2
Step 2
Alternative to claim is π1 β₯ π2
Step 3
The null and alternative hypothesis are:
π»0 : π1 = π2 .
π»π : π1 < π2 (Claim)
Step 4
significance level is ο‘ = 0.05
Step 5
π₯1 = 120, π1 = 200.
π₯2 = 150, π2 = 200.
πΜ =
zο½
π₯1 +π₯2
π1 +π2
= 0.675, πΜ = 0.325
ο¨pΜ1 ο­ pΜ2 ο©ο­ ο¨p1 ο­ p2 ο©
pq
pq
ο«
n1
n2
120 150
β
200
200
π§=
= β3.2
0.675
×
0.325
0.675
×
0.325
β
+
200
200
Page 8 of 10
So the P- value is 0.0007.
Step 6
The P-value of 0.0007 is less than the significance level of ο‘ = 0.05, we
reject the null hypothesis of π1 = π2 .
Because we reject the null hypothesis, we conclude that there is sufficient
evidence to support the claim.
Q. 5 & 6.Use the following information to answer Questions 5 and 6:
Given the following data of two independent samples of normally
distributed populations
Data
Population 1
Population 2
n
23
13
43
41
4.5
5.1
π₯Μ
s
5. Test the claim that π1 β  π2 at the Ξ±=0.05 level of significance
6. Construct a 95% confidence interval about π1 β π2 .
Solution(5) : two samples are independent and
π₯1 = 43, π1 = 23, π 1 = 4.5.
ΜΜΜ
π₯2 = 41, π2 = 13, π 2 = 5.1.
ΜΜΜ
Consider
π1 : mean of population 1
π2 : mean of population 2
Step 1
The claim is π1 β  π2
Step 2
Alternative to claim is π1 = π2
Step 3
The null and alternative hypothesis are:
π»0 : π1 = π2 .
π»π : π1 β  π2 (Claim)
Step 4
Significance level is ο‘ = 0.05
Page 9 of 10
Step 5
π₯1 = 43, π1 = 23, π 1 = 4.5.
ΜΜΜ
π₯2 = 41, π2 = 13, π 2 = 5.1.
ΜΜΜ
π‘=
(π₯
ΜΜΜ1 β ΜΜΜ)
π₯2 β (π1 β π2 )
π 2 π 2
β 1+ 2
π1 π2
π‘=
43 β 41
β4.5 × 4.5 + 5.1 × 5.1
23
13
= 1.178
Step 6:
The t- value at πΌ/2 = 0.025 and df=n-1=13-1=12 is ±2.179.
Because the test statistic does not fall within the critical region, fail to
reject the null hypothesis: π1 = π2 .
Solution(6): Margin of error (E) is given by
πΈ=
π 12
β
πΌ
π‘ β2
π1
π 22
+
π2
Thus
4.5 × 4.5 5.1 × 5.1
πΈ = 2.179β
+
= 2.179 × 1.697 = 3.699
23
13
Confidence interval is given by
(π₯
ΜΜΜ1 β ΜΜΜ)
π₯2 β πΈ < (π1 β π2 ) < (π₯
ΜΜΜ1 β ΜΜΜ)
π₯2 + πΈ
(43 β 41) β 3.699 < (π1 β π2 ) < (43 β 41) + 3.699
β1.699 < (π1 β π2 ) < 5.699
Page 10 of 10
```
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