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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
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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
= 0.271 Z = 3.34
Z
0.05/2 : 1.96
Reject the null hypothesis since there is enough significant evidence to conclude that there’s a difference in patients experiencing pain relief. Question 7
Null hypothesis H
O : μ = 0
H
1 : μ not equal to 0
Level of significance α = 0.05 α/2 = 0.025
Degree of freedom = n-1 = 7-1 = 6
T value according to the table = 0.025
N =7 xi
x
i
– x
(x
i
– x)^2
122-120
2
1.14
1.2996
142-145
-3
-3.86
14.8996
135-130
5
4.14
17.1396
158-160
-2
-2.86
8.1796
155-152
3
2.14
4.5796
140-143
3
-3.86
14.8996
130-126
4
3.14
9,8596
= 70.8572
Total x
i = 6
Mean = 0.86
Standard deviation = n= 7 sd = 3.4365
T = 0.86 -0 / (3.4365/ square root 7) = 0.662
We fail to reject the null hypothesis as the t test value is less than the critical value. Question 8 a
x1 = 120.2. x2 = 131.4. α = 0.05 critical value = 1.96
S1 = 15.4. s2 = 18.9
n1 = 100. n2 = 100
null hypotheses and alternate hypotheses H
O
: μ1 = μ2
H
1 : μ1 not equal to μ2
Test statistics sp
2
= (n1-1)*s1
2
+(n2-1)*s2
2 / n1+n2-2 = 297.185
SE = square root (sp
2
*(1/n1 +1/n2) = 2.438
Test statistics = 120.2 – 131.4 / 2.438 = -4.5939
Hence -4.59 < 1.96 we reject the null hypotheses.
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Question 9
Null hypothesis: H
O
: p1=p2=p3
Alternate hypothesis = p1 not equal to p2 not equal to p3
Site 1
Site 2
Site 3
total
hypertensive
10
14
12
36
Non hypertensive
68
56
40
164
total
78
70
52
200
Expected frequencies: Site 1
Site 2
Site 3
hypertensive
14.04
12.6
9.36
Non hypertensive
63.96
57.40
42.64
Observed frequencies
Expected frequencies
Observed frequency – expected frequency (O-E) ^2
(O-E) ^2/E
10
14.04
-4.04
16.3216
1.162
14
12.60
1.4
1.96
0.155
12
9.36
2.64
6.9696
0.744
68
63.96
4.04
16.3216
0.255
56
57.40
-1.4
1.96
0.034
40
42.64
-2.64
6.9696
0.163
= 2.515
Test statistics = 2.515
Degree of freedom = (r-1) (c-1)
(2-1) (3-1) = 1*2 = 2
P value 1- 0.7157 = 0.2843 0.2843 > 0.05
Null hypothesis is not rejected. there is not enough evidence to claim that the two variables are dependent at 5% significant level. Question 10
H
O : P1 = P2
H
1 : P1 not equal P2. Level of significance = 0.05. 1.96
P1= x1/n1 = 38/100 = 0.38
P2 = x2/n2 = 21/100 = 0.21
P = 38 +21/200 = 0.295
Z = 0.38 – 0.21/ square root (0.295 (1-0.295) (1/100+1/100) = 0.17/0.0644 = 2.636
We reject the hull hypothesis as 2.636 is greater than 1.96.
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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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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
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the solution is not clearly written here
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Question 6
The data shown in Table 3 are and R values for 24 samples of size n = 5 taken from a process producing
bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals
recorded (i.e., 34.5 should be 0.50345). Perform a process capability analysis using and R charts for the data
from Table 3.
Table 3: Data for Question 6
Sample Number
R | Sample Number I
Ꭱ
1
34.5
3
13
35.4 8
2
34.2
4
14
34.0 6
∞∞
31.6
15
37.1 5
4
31.5 4
16
34.9 7
56789
35.0
5
17
33.5
4
34.1 6
18
31.7 3
32.6 4
19
34.0
8
33.8 3
20
35.1 4
34.8 7
10
33.6
8
11
31.9 3
12
38.6
9
2232
21
33.7
2
32.8 1
33.5
24
34.2 2
32
3
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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 = ______
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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?
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Please answer these two short questions
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3. solve for
s2 =
t =
t >
t <
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E 3 of 12 Questions
Completed 7 out of 12
Resources
Submit All
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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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Question3: If the researcher in Question 1 increases the sample size to 64, calculate the 90% CI for the average GPA of college students with the new sample size.
What is the standard error?
What is the critical z or t score?
What is the CI?
Interpret the 90% CI.
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14) part B
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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
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Steps 4,7 and 2B please.
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Question 2
Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples
of five parts each are collected, and the compressive strengths (in psi) are shown in Table 2.
Table 2: Strength Data for Question 2
Sample Number
x1
x2
23
x4
x5
R
1
83.0
2
88.6 78.3 78.8
3
85.7
75.8
84.3
81.2 78.7 75.7 77.0
71.0 84.2
81.0
79.1
7.3
80.2 17.6
75.2
80.4
10.4
4
80.8
74.4
82.5
74.1 75.7 77.5
8.4
5
83.4
78.4
82.6 78.2
78.9
80.3
5.2
File Preview
6
75.3
79.9
87.3 89.7
81.8
82.8
14.5
7
74.5
78.0 80.8
73.4
79.7
77.3
7.4
8
79.2
84.4 81.5 86.0
74.5
81.1
11.4
9
80.5
86.2
76.2 64.1
80.2
81.4
9.9
10
75.7
75.2
71.1 82.1
74.3
75.7
10.9
11
80.0 81.5
78.4 73.8
78.1
78.4
7.7
12
80.6
81.8
79.3
73.8
81.7 79.4
8.0
13
82.7
81.3
79.1
82.0 79.5 80.9
3.6
14
79.2
74.9
78.6 77.7
75.3
77.1
4.3
15
85.5 82.1
82.8 73.4
71.7
79.1
13.8
16
78.8 79.6
80.2 79.1
80.8 79.7
2.0
17
82.1
78.2
18
84.5
76.9
75.5
83.5 81.2
19
79.0 77.8
20
84.5
73.1
78.2 82.1
79.2 81.1 7.6
81.2 84.4 81.6 80.8…
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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.
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Attached is the image of my homework. I was wondering how I know if it's directional or non directional hypothesis based on the problem. Additionally, I was wondering how to calculate df since this is t test for two independent samples. thank you!
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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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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
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Match each step on the left with the procedure at
that step on the right.
Step
1
3
Procedure
a. calculate test statistics
b. compare p-value with alpha
c. write null and alternative hypotheses
d. identify test type
e. write conclusion
f. reject or fail to reject null hypothesis
g. calculate p-value
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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
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Question 44
Given the following hypothesis test:
Ho: HxS20
Ha: Hx>20
and this information:
n= 36 , X = 24.6, Sx = 12 , 0x is unknown
%3D
What would be the value of the test statistic for this test?
O 2.3
O 0.38
O-2.3
O -0.38
O none of these
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Thu 9:45 PM
BExtra Credit Sx
b My Questions
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Unit 7.3 Condu
G Nancy, an eco
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X
bconline.broward.edu/d2l/le/content/407158/viewContent/10432138/View
ot
9 PM
Make a Conclusion and Interpret the Results of a One-Mean Hypothesis Test (Population Standard Deviation Known) Using the P-Value Approach
Question
ot
Nancy, an economist, would like to make the claim that the average weekly grocery budget for households in a small U.S.
city is less than $253. Nancy samples 18 households in the same city and obtains a sample mean of $229.60 spent on
18 PM
groceries per week.
At the 2.5% significance level, should Nancy reject or fail to reject the null hypothesis given the sample data below?
9 PM
Ho : u = $253 per week; Ha :u < $253 per week
0.025 (significance level)
a
ot
4 PM
test statistic = -0.75
Use the graph below to select the type of test (left-, right-, or two-tailed)
O PM
Then set the a and the test…
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urgently please answer question 2 and 3
QUESTION 2
An experiment was conducted into car tyre wear by the department of Materials Science at Lowlands University, which it has been asked to carry out by a tyre company.
The tyre company hascreated a new compound which it hopes will make tyres last much longer. Various levels of the compound were mixed in with rubber to make tyres. Apart from that, all the tyres were identical, with 16mm of tread when new. One hundred motorists chosen at random were given new tyres, and split into ten groups. After each motorist had completed 10,000 miles on the new tyres, the level of tread remaining was measured. The data recording the experiment is set out in the table below.
You have been asked to use correlation analysis to see if there is a relationship between the percentage of the new compound and the wear on the tyres.
Identify appropriate x and y variables, stating clearly which variable is the “x” variable and which is the “y” variable…
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9:27 PM Thu Apr 6
<
no
BEE
a
Child
1
2
3
4
6
5
17. Psychostimulant drugs (such as Ritalin) help children with ADHD focus better on classroom material. But a
researcher wonders whether Ritalin might also influence children's ability to engage in productive social
interactions. She selects 6 children who are newly diagnosed with ADHD. During one week, participants receive
Ritalin, and during another week, the same participants receive a placebo. Their classroom teacher rates their
social skill at the end of each week (data are shown below; higher scores indicate better social interactions).
Consider the appropriate test for this question and answer Parts a-d. Note: you do NOT need to calculate
means, standard deviations or the test statistic!!
Homework 6
Mean
SD
Social score for
Ritalin week
(points)
23
12
15
20
21
19
18.33 points.
4.08 pts
Untitled (Draft)
Social score for
Placebo week
(points)
18
18
5
16
22
11
15.00 points
6.07 pts
A) (b) What is the null hypothesis in symbols?
T
n…
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