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Week 9 Question 1 Step 1Null hypothesis H
O
: μ SBP = 120 mmhg
Alternate hypotheses H
1
: placebo SBP > μ SBP
Level of significance 0.05
Step 2 Because the population size is greater than 30 we will use the z method
Step 3 we will reject the hypothesis if μ = 120 mmhg
Step 4 Sx=
s
√
n
Sx = 18.9/ √
100
= 1.89
We find z
Z = (x – μ )/Sx. = 131.4
−
120
1.89
= 6.03
Step 5 We reject the null hypothesis as 6.03 > 1.654
Question 2 :
Step 1 H
O
: vitamin level of cancer patients = 3.00 = μ
vitamin
H
1 : population μ
vitamin > cancer patient vitamin level
Step 2 : using t methods because the sample size n is less than 30
Step 3: we will reject the null hypothesis is the value is less than or eqal to -1.645
Step 4: Sx=
s
√
n
= 0.15/ √
20
= 0.034
t= x
−
μ
Sx
= 2.41
−
3
0.034
= -17.35
we will reject the null hypothesis as the value is less than -1.645. the vitamin levels of cancer patients are not the same as the vitamin level of population. Question 3:
Step 1 Null hypothesis H
O
: placebo data side effect = 0.05
Alternate hypothesis H
1 : placebo data side effect isn’t equal to 0.05
Step 2 : as the value of n is greater than 30 we will use z method Step 3: reject the null hypothesis if greater or equal to 1.960 Step 4 Z = p
−
p
0
√
p
O
(
1
−
p
0
)
n
= 1.376
Step 5 We fail to reject the null hypothesis as the value is less than 1.960. Question 4
Step 1: Null hypothesis H
O
: P1=0.20 P2= 0.20
P3=0.20
P4 =0.20
Alternate hypothesis H
1 : that the values of the null hypothesis isn’t true
Level of significance 5% = 0.05
Step 2 : as its categorial we use the chi square method
Step 3: if the value of x
2 is greater than 9.49 we will reject the null hypothesis Step 4: X
2 = (
(
O
−
E
)
2
)
E
= 206.9
STEP 5: we reject the null hypothesis as the value is greater than 9.49
Week 10 Question 1 H
O : The new compound treatment, and the extent of wound healing is independent of each other.
H
1 : The new compound treatment, and the extent of wound healing is dependent of each other.
Level of significance = 5% α = 0.05
Treatment
0-25
26-50
51-75
76-100
New compound n = 125
15
37
32
41
Placebo n = 125
36
45
34
10
total
15+36 = 51
37+45 =82
32+34 = 66
41+10 = 51
Frequency table treatment
0-25
25-50
51-75
76-100
New compound
(125*51)/250
= 25.5
(125*82)/250 = 41
(125*66)/250 = 33
(125*51)/250
= 25.5
Placebo
(125*51)/250
= 25.5
(125*82)/250 = 41
(125*66)/250 = 33
(125*51)/250
= 25.5
E
O-E
(O-E)^2
(O-E)^2/E
15
25.5
-10.50
110.25
4.3235
37
41
-4
16
0.3902
32
33
-1
1
0.0303
41
25.5
15.50
240.25
9.4216
36
25.5
10.5
110.25
4.3236
45
41
4
16
0.3902
34
33
1
1
0.0303
10
25.5
-15.50
240.25
9.4216
Total
= 28.33
Test statistics
= 28.33
DF = k-1 (4-1) (2-1)
3*1 =3
According to the table: 7.8177
There are enough evidence to reject the null hypothesis treatment, since the test statistics 28.331 is greater than critical value.
Question 6 P1 be the proportion of patients who get relied from new medication.
P2 be the proportion of patients who get relied from placebo. H
O : P1 = P2
H
1 : P1 not equal P2. Sample proportion of pain relief for two treatments = x1/n1
= 44/44+76 = 0.367
Pooled estimate of the sample proportion = x1 + x2 / n1 + n2
= 44 +21 / 120 + 120
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Related Questions
Referring to Table 1, which of the following is an appropriate alternative hypothesis?
Question 25 options:
1)
H1 : μI - μII = 0
2)
H1 : μI - μII < 0
3)
H1 : μI - μII > 0
4)
H1 : μI - μII ≠ 0
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question 3 part 3
the equation for the problem is picture number 2
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If H1: μ1 – μ0 > 0 and α = 0.05, what effect size seems to interest me if I wish to achieve power =0.05 with a sample size of 44?
The answer is: 0.5
Please provide a solution to this problem and explain the answer
arrow_forward
If H1: μ1 – μ0 > 0 and α = 0.05, what effect size seems to interest me if I wish to achieve power =0.8 with a sample size of 13?
a. 0.5
b. 0.1
c. 0.6
d. 0.8
arrow_forward
PRINTER VERSION
Chapter 09, Section 9.2, Problem 014c
( BACK
NEXT
Consider Ho: µ = 38 versus H1: µ> 38. A random sample of 35 observations taken from this population produced a sample mean of 40.29. The population is normally distributed with o = 7.2.
Calculate the p-value. Round your answer to four decimal places.
p =
the tolerance is +/-2%
Question Attempts: 0 of 2 used
SAVE FOR LATER
SUBMIT ANSWER
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MacBook Air
888
F12
F11
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F10
F8
F9
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F4
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F2
23
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del
%3D
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8
9
2
4
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the solution is not clearly written here
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Scenario 2
To measure whether test performance can be predicted based on one's anxiety, a researcher asked his
students to come to the lab 15 minutes before they were to take an exam in his class. The researcher measured
the students' heart rates and then matched these scores with their exam performance after they had taken the
exam. Use the data below and SPSS to test whether heart rate can predict test performance in the population.
田
Student
Heart Rate
Exam Score
A
76
78
B
81
68
60
88
D
65
80
E
80
90
F
66
68
82
60
H
71
95
I
66
84
J
75
75
K
80
62
L
76
51
M
77
63
N
79
71
7. In this specific scenario, are the two variables of interest? Which is the IV and which is the DV? What
is the scale of measurement for each variable?
Page 3 of 7
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Speed w/
ordinary
shoes
Speed w/
special
shoes
Runner
INSTRUCTIONS:
1. State the null
and alternative
hypotheses.
A
12.1 mi/h
12.5 mi/h
2. State the level of
14.4 mi/h
15.0 mi/h
significance and
critical value.
C
11.6 mi/h
11.4 mi/h
3. Calculate the
11.2 mi/h
11.6 mi/h
ttest.
E
12.4 mi/h
11.8 mi/h
4. Interpret the
result.
F
11.6 mi/h
12.0 mi/h
5. State conclusion.
G
15.0 mi/h
14.9 mi/h
arrow_forward
ht than
ollow
Problem 2
Tellowing
A study is conducted on the effects of two drugs (simply called Drug A and Drug B) upon
hyperactivity in laboratory rats. Two random sample of rats are used for the study - one sample
receiving Drug A, the other receiving Drug B. After two weeks, the rats are measured on
hyperactivity with the following results:
Drug A: X, = 75.6 S1 = 3.5, n, = 18
Drug B: X2 = 72.8 S2 = 3.2 n2 = 24
1) State the null hypothesis and the alternative hypothesis.
2) Find the critical value. Use a = 0.05
3) Find the standardized t test.
4) Find the p-value
5) Decide whether to reject or fail to reject the null hypothesis.
6) Interpret the result.
arrow_forward
Scenario: 200 people were asked, “Who is your favorite superhero?” Below are the data. Test the null hypothesis that the population frequencies for each category are equal. α= .05.
Iron Man
Black Panther
Wonder Woman
Spiderman
fo = 40
fo = 60
fo =55
fo = 45
fe =
fe =
fe =
fe =
degrees of freedom, df = ______, and the Chi Square critical boundary = ______
arrow_forward
Step 1 of 3 :
State the null and alternative hypotheses for the test. Fill in the blank below.
H0: μd=0
Ha: μd⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0
Step 2:
What is the test statistic?
Step 3:
Do we reject the null hypothesis? Is there sufficient or insufficient data?
arrow_forward
No need for explanation
arrow_forward
Ho : H= 1.87
H1 : u tcrit = 1.645, then reject Ho; otherwise fail to reject Ho.
Oa = 0.05; one-tailed; if tobt > tcrit= 1.761, then reject Ho; otherwise fail to reject Ho.
Oa= 0.05; one-tailed; if zobt < Zcrit = -1.645, then reject Ho; otherwise fail to reject Ho.
%3D
α.
QUESTION 20
For 6 months we worked with a groun of 15 severelv mentally handicapped individuals in an attempt to teach them self-care through imitation. For a second 6-
Save All Ar
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
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Please answer these two short questions
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Please help with this statics question, not sure if my answers are correct. Would you please help solve? Thanks
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E 3 of 12 Questions
Completed 7 out of 12
Resources
Submit All
Press esc to exit full screen
1 Question
Question 3 of 12
>
An analyst at a state's environmental quality commission has just finished processing salt concentration tests on 60 wells in
2 Question
Washington County to represent the more than 1,000 wells across the state. Water with salinity concentration less than 0.05% is
considered fresh.
3 Question
0.03 on this sample of wells to determine if the
overall salt concentration of the state's water supply exceeds 0.05%. The shape of the sample data is approximately symmetric
The analyst decides to run a one-sample t test at a significance level of a =
4 Question
with no outliers, and its statistics are
5 Question
x = 0.048%
Sx =
:0.0366%
6 Question
P-value
:0.0166
7 Question
where x is the sample mean, sz is the sample standard deviation, and the P-value is the probability that results from the t test.
The analyst decides that there is enough evidence to conclude that the…
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#2
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ASSESSMENT OF PERFORIMANLE / PRODUCT
Direction: Apply the steps that you learned in hypothesis testing to the given Problem below. Write
your answer on a separate sheet of paper.
Problem :
Consider a study wherein the effectiveness of banana leaves as an organic cooked rice
Wrapper was compared to that of alumimum foil The number of bacterial colonies in the two
kinds of food wrappers was measured after 12 hours at room temperature. The ev is 28, £=
0.01. The result of computad value is 0.8.
Hvnothesis:
Ho:
2.
Ha:
3. Level of Sienificance:
4. Statistical Tool:
5. Computation Results:
6. Decision Rule:
7. Internretation:
arrow_forward
Question 44
For a population with u = 50 and o = 10, a score of X 55 corresponds to z = +0.50.
True
False
Question 45
arrow_forward
Question 2
You are developing a new analytical method for the determination of blood glucose content.
You want to ascertain whether your method differs significantly from a standard method for
determining a range of sample concentrations expected to be found in the routine laboratory.
It has been ascertained that the two methods have comparable precision. Following Table 2.0
are two sets of results for a number of individual samples:
Sample
b
с
d
e
f
Table 2.0
Your method
(mg/L)
10.2
12.7
8.6
17.5
11.2
11.5
Standard method
(mg/L)
10.5
11.9
8.7
16.9
10.9
11.1
(a) State the null and alternative hypotheses.
(b) Determine whether there are differences in the two methods at the 95% and 99%
confidence levels.
arrow_forward
Steps 4,7 and 2B please.
arrow_forward
QUESTION 13
Suppose dfErr = 16, SST = 1917, and Se = 3.9 for the simple regression ý = bo + b1(x).
What is r2?
a. 0.8731
b. None of the answers is correct
c. 0.8874
d. 0.8811
e. 0.8651
QUESTION 14
New
Last Year
Comple Cinall
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
MacE
20
D00 FA
F1
F3
F5
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QUESTION 4
A drug company is testing to see if their new drug does lower total cholesterol. The used a population who had a total cholesterol reading of 32
5 mg/dL prior to using the drug.
Ho: H = 325
Ha: H<325
Suppose the results of their testing lead to rejection of the null hypothesis. Classify that conclusion as: correct, Type I error or Type II error, if in
fact the total cholesterol levels have decreased.
arrow_forward
(No. 1)
Given: μ = 48, s = 17, n=25 and x = 65
STATISTICAL TOOL: T-TEST: ONE SAMPLE MEAN
Formula:
Solution:
(No. 2)
Given: x₁ = 7; x₂= 5; 5₁² = 2; 5₂² = 4; n₁ = 10; n₂ = 10
STATISTICAL TOOL: T-TEST: TWO SAMPLE MEANS
Formula:
Solution:
arrow_forward
Solve the problem.
From the TI-84 graphing calculator screenshots below,
choose the screenshot whose shaded area correctly
depicts the following hypothesis test results:
Ho: p = 0.25, Ha: p > 0.25, a = 0.05, z = 2.01, p-value
= 0.022
O
O
Ar&q=.420349
10w=.201
lup=10 01
AP44=.022216
low=2.01
up=10
Y
S
arrow_forward
9
arrow_forward
E.
D.
Species Sum of Sample Values
A
Σi Yai = 42.3
B
C
Σi-1 YBi = 44.1
ΣΥΒ
Yci = 48.6
Calculate the sum of squares for treatment in a one-way ANOVA with
species as the factor.
Jade asked her colleague to perform a two-way ANOVA on the same
sample data with food type and species as the two factors. Her colleague
performed the two-way ANOVA (assuming any required assumptions were
satisfied) but unfortunately misplaced the results. All that he could remember was
that there was an interaction between food type and species at the 5%
significance level but not at the 2.5% significance level. Calculate the two values
between which the sum of squares for interaction in this two-way ANOVA must lie.
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G
Mc
Graw
Hill
F5
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5
D
T
F6
6
Y
&
67°
DELL
7
Terms of Use Minimum Requirements Platform Status Center
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Answer
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Question 5/9
56 %
If us0.34 is the stated null hypothesis, what is the alternative hypothesis?
Ο μ0.34
Ο μ20.34
Type your answer here:
arrow_forward
Exhibit 9-31
n = 64
x= 32.8
o=15
Ho: μ = 30
H₂:30
Refer to Exhibit 9-31. If the test is done at a 5% level of significance, the null hypothesis should
not be rejected
None of the answers is correct.
Not enough information is given to answer this question.
be rejected
arrow_forward
QUESTION 10
Find the critical value or values of y2 based on the given information.
H1: o > 26.1
n = 9
a = 0.01
1.646
21.666
O 20.090
O 2.088
arrow_forward
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- If H1: μ1 – μ0 > 0 and α = 0.05, what effect size seems to interest me if I wish to achieve power =0.8 with a sample size of 13? a. 0.5 b. 0.1 c. 0.6 d. 0.8arrow_forwardPRINTER VERSION Chapter 09, Section 9.2, Problem 014c ( BACK NEXT Consider Ho: µ = 38 versus H1: µ> 38. A random sample of 35 observations taken from this population produced a sample mean of 40.29. The population is normally distributed with o = 7.2. Calculate the p-value. Round your answer to four decimal places. p = the tolerance is +/-2% Question Attempts: 0 of 2 used SAVE FOR LATER SUBMIT ANSWER powered by MapleNet on y Study Version 4.24.20.1 nt I Privacy Policy I 2000-2020 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc. MacBook Air 888 F12 F11 80 F10 F8 F9 F6 F7 F4 F5 F1 F2 23 2$ del %3D 7 8 9 2 4arrow_forwardthe solution is not clearly written herearrow_forward
- Scenario 2 To measure whether test performance can be predicted based on one's anxiety, a researcher asked his students to come to the lab 15 minutes before they were to take an exam in his class. The researcher measured the students' heart rates and then matched these scores with their exam performance after they had taken the exam. Use the data below and SPSS to test whether heart rate can predict test performance in the population. 田 Student Heart Rate Exam Score A 76 78 B 81 68 60 88 D 65 80 E 80 90 F 66 68 82 60 H 71 95 I 66 84 J 75 75 K 80 62 L 76 51 M 77 63 N 79 71 7. In this specific scenario, are the two variables of interest? Which is the IV and which is the DV? What is the scale of measurement for each variable? Page 3 of 7arrow_forwardSpeed w/ ordinary shoes Speed w/ special shoes Runner INSTRUCTIONS: 1. State the null and alternative hypotheses. A 12.1 mi/h 12.5 mi/h 2. State the level of 14.4 mi/h 15.0 mi/h significance and critical value. C 11.6 mi/h 11.4 mi/h 3. Calculate the 11.2 mi/h 11.6 mi/h ttest. E 12.4 mi/h 11.8 mi/h 4. Interpret the result. F 11.6 mi/h 12.0 mi/h 5. State conclusion. G 15.0 mi/h 14.9 mi/harrow_forwardht than ollow Problem 2 Tellowing A study is conducted on the effects of two drugs (simply called Drug A and Drug B) upon hyperactivity in laboratory rats. Two random sample of rats are used for the study - one sample receiving Drug A, the other receiving Drug B. After two weeks, the rats are measured on hyperactivity with the following results: Drug A: X, = 75.6 S1 = 3.5, n, = 18 Drug B: X2 = 72.8 S2 = 3.2 n2 = 24 1) State the null hypothesis and the alternative hypothesis. 2) Find the critical value. Use a = 0.05 3) Find the standardized t test. 4) Find the p-value 5) Decide whether to reject or fail to reject the null hypothesis. 6) Interpret the result.arrow_forward
- Scenario: 200 people were asked, “Who is your favorite superhero?” Below are the data. Test the null hypothesis that the population frequencies for each category are equal. α= .05. Iron Man Black Panther Wonder Woman Spiderman fo = 40 fo = 60 fo =55 fo = 45 fe = fe = fe = fe = degrees of freedom, df = ______, and the Chi Square critical boundary = ______arrow_forwardStep 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μd=0 Ha: μd⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0 Step 2: What is the test statistic? Step 3: Do we reject the null hypothesis? Is there sufficient or insufficient data?arrow_forwardNo need for explanationarrow_forward
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