- Statistics assignment
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- May 13, 2015 | Wednesday 01:09 pm | 2 months ago
Question 1 (9 marks in total)
Cans of Fizzy Cola are believed to have contents that are Normally distributed with a mean of
305 ml and a standard deviation of 10 ml.
(a) If a random sample of 25 cans is selected from a day’s production, what is the probability
that the 12th can contains more than 300 ml? (2 marks)
(b) If the average of the contents of the 25 cans is calculated, what is
i. the standard error of this average? (1 mark)
ii. the probability that this average is greater than 300 ml? (3 marks)
(c) What is the probability that at least 20 of the 25 cans have contents greater than 300 ml?
(3 marks)
Marking Criteria
Marks 0 1 2 3
1(a) Incorrect answer and
no working
Correct answer but
working has errors
Correct answer with
correct working
1(b)i. Answer is incorrect Answer is correct
1(b)ii.
and
1(c)
Solution is missing or
incorrect
One part of precess
is correct, but rest is
missing or wrong
Process attempted but
impaired by a minor er-
ror, or correct
nal an-
swer with inadequate
working
Correct
nal answer
with working (interme-
diate calculations or
output from software)
3
STA201/401 201530 Assignment 2
Question 2 (15 marks in total)
Suppose that, in 10 years of use throughout Australia, one particular animal vaccine has been
successful in treating an animal disease in 66% of cases (i.e., on average, 66% of animals have
recovered), and this percentage has become accepted as the industry `standard’.
(a) JaneyWerakso has 64 animals su
ering from this disease. If she administers the vaccine
in the approved manner, what is the standard error of the proportion of animals that will
recover? (2 marks)
(b) After treatment, Janey
nds that 39 of her animals recover from the disease. What
proportion of Janey’s animals recover from this disease? (1 mark)
(c) What is the probability of getting this proportion (or less) of recoveries from 64 animals
if the vaccine really does have a 66% recovery rate? (3 marks)
(d) If p represents the true probability that an animal recovers from the disease when the
vaccine is administered, use the data from Janey’s experiment to test H0 : p = 0:66 vs
H1 : p 6= 0:66 at the 5% level of signi
cance. (5 marks)
(e) Janey is pretty annoyed that less than 66% of her animals recovered, and wants to know
why we didn’t test H0 : p = 0:66 vs H1 : p < 0:66 because, after all, less than 66% recovered.
Is this a valid thing to do? Answer `yes’ or `no’, and justify your answer. (2 marks)
(f) In the light of Janey’s concerns, would it be valid to take another large sample of animals
and test H0 : p = 0:66 vs H1 : p < 0:66? Justify your answer. (2 marks)
Marking Criteria
Marks 0 1 2 3
2(a) Incorrect answer and
no working
Correct answer but
working has errors
Correct answer with
correct working
2(b) Answer is incorrect Answer is correct
2(c) Incorrect answer and
no working
Incorrect answer, but
some sensible working
Correct answer, but
working incorrect or in-
complete
Correct answer with
correct working
2(d) (See below)
2(e)
and
2(f)
Answer (Y/N) incor-
rect, and no justi
ca-
tion
Answer correct, but
justi
cationinade-
quate
Correct answer, and
justi
cation that is cor-
rect and clear
Marks 0 1
2(d) Test statistic and
samplingdistribu-
tion
Statistic and distribution either
missing or incorrect
Statistic and distribution both
correct
Rejection criterion Criterion missing or incorrect Criterion correctly stated
Calculations (soft-
ware or manually)
Calculations wrong or only par-
tially correct
Calculations correct
Decision Decision missing or not justi
ed Decision correctly stated and
justi
ed
Conclusion Missing, or not in context of prob-
lem
Statement in English in context of
problem
4
STA201/401 201530 Assignment 2
Question 3 (12 marks in total)
Suppose that a reputable survey agency is engaged to estimate the proportion of Australian
electors who believe that, to reduce carbon emissions, a nuclear power plant should be built
somewhere in Australia.
(a) Under the assumption that everyone who is contacted gives a reply of `In favour’ or `Op-
posed/Don’t Know’, how large a sample must be taken if the estimate obtained from the
survey is to have a margin of error of at most 2%? Use a 95% con
dence interval in
calculating the margin of error, and assume that absolutely nothing is known about the
proportion of Australians in favour. (3 marks)
(b) Repeat (a) if it is believed that the proportion of Australian electors in favour is not more
than 30%. (2 marks)
(c) Repeat (a) if it is thought that 20% of people contacted will refuse to answer, thereby
providing no information. (Make the assumption that these people think like the other
80%, but just don’t want to answer.) (2 marks)
(d) Suppose that a random sample of 1000 Australian electors is asked whether they believe
that, to reduce carbon emissions, a nuclear power plant should be built somewhere in
Australia. After eliminating the 221 people who refuse to answer, 275 of the remainder say
that they support building a nuclear power plant. Calculate a 95% con
dence interval for
the true proportion, p, of Australian electors who believe that, to reduce carbon emissions,
a nuclear power plant should be built somewhere in Australia. (3 marks)
(e) Use the con
dence interval obtained in (d) to test H0 : p = 0:30 vs H1 : p 6= 0:30 at the 5%
level of signi
cance. You do not have to carry out the formal steps of hypothesis testing,
but your answer must say whether or not you reject H0, and give reasons for your answer.
(2 marks)
Marking Criteria
Marks 0 1 2 3
3(a) Incorrect answer and
no working
Incorrect answer but
partly correct working
Correct answer with
partly correct working
Correct answer, with
working that is clear
and correct
3(b)
and (c)
Incorrect answer and
no working
Incorrect answer but
partly correct working
Correct answer, with
working that is clear
and correct
3(d) Incorrect answer and
no working
Correct working, but
no recognition of non-
responses
Correct answer with-
out complete explana-
tion
Correct answer, with
working that is clear
and correct
3(e) Incorrect answer and
no working
Correct answer, but in-
complete explanation
Correct answer, with
working that is clear
and correct
5
STA201/401 201530 Assignment 2
Question 4 (12 marks in total)
A researcher is investigating the e
ectiveness of a new diet that is being considered for raising
chickens. He wants to measure the weight gained by chicks on this diet in the 14 days between
the 7th and 21st days of life. On a given date, a random sample of 25 7-days-old chicks is
selected, and their weights are measured then and again on the 21st day. The weight gains (in
kg) are recorded in the
le chicken weights.csv in the folder in Interact2 where this Assignment
is located.
The researcher wants to estimate the mean weight gained (in kg), , by chicks on this diet.
He asks you to calculate a 95% con
dence interval for the value of , and tells you to use the
t-distribution.
(a) What assumptions are required to be met for the use of the t-distribution? Do they seem
valid here? Justify your answer. (3 marks)
(b) Irrespective of your answer to (a), calculate a 95% con
dence interval for . (3 marks)
(c) The mean weight gain over 14 days under a standard diet is 0.85 kg. (This
gure has been
established over several years, and can be regarded as a known constant.) Test at the 1%
signi
cance level whether the new diet has a mean weight gain that is greater than 0.85
kg. Give all the steps of the hypothesis testing process. However, you should use the R
Commander to perform the calculations. (6 marks)
Marking Criteria
Marks 0 1 2 3
4(a) All assumptions incor-
rect
All assumptions cor-
rect, but their validity
not assessed
All assumptions cor-
rect, but the validity of
only one assessed
All assumptions cor-
rect and their validity
assessed
4(b) Answer and working
incorrect
Answer wrong but
working partly correct
Answer correct but
working incompletely
described
Answer correct and
working correct and
clear
Marks 0 1
4(c) H0, H1 and
Hypotheses missing or incorrectly
stated
Hypotheses correctly stated and
symbols explained,
speci
ed
Test statistic and
samplingdistribu-
tion
Statistic and distribution either
missing or incorrect
Statistic and distribution both
correct
Rejection criterion Criterion missing or incorrect Criterion correctly stated
Calculations (soft-
ware or manually)
Calculations wrong or only par-
tially correct
Calculations correct
Decision Decision missing or not justi
ed Decision correctly stated and
justi
ed
Conclusion Missing, or not in context of prob-
lem
Statement in English in context of
problem
6
STA201/401 201530 Assignment 2
Question 5 (12 marks in total)
(This is based on a question from Daniel, W.W. and Cross, C.L. (2013). Biostatistics: A
Foundation for Analysis in the Health Sciences. (10th edn) Hoboken, NJ: Wiley.)
Researchers investigated the e
ect of broadband ultraviolet B (UVB) therapy and a topical
cream used together on areas of psoriasis (a skin complaint). One of the outcome variables was
the Psoriasis Area and Severity Index (PASI). The Table that appears below contains the PASI
scores for 20 patients who were assessed at baseline and after eight treatments.
Subject 1 2 3 4 5 6 7 8 9 10
Baseline 5.9 7.6 12.8 16.5 6.1 14.4 6.6 5.4 9.6 11.6
After 8 Treatments 5.2 12.2 4.6 4.0 0.4 3.8 1.2 3.1 3.5 4.9
Subject 11 12 13 14 15 16 17 18 19 20
Baseline 11.1 15.6 6.9 15.2 21.0 5.9 10.0 12.2 20.2 6.2
After 8 Treatments 11.1 8.4 5.8 5.0 6.4 0.0 2.7 5.1 4.8 4.2
The original question asked `Do these data provide evidence, at the .01 level of signi
cance, to
indicate that the combination therapy reduces PASI scores?’
(a) State the null and alternative hypotheses that you would test in order to answer this
question. If you use symbols other than H0 and H1 (and `=’, etc) in your statement, you
must explain what those symbols represent. (2 marks)
(b) If the data satisfy the appropriate statistical assumptions, then these hypotheses can be
tested by using a t-test. Would you use a `Welch Two Sample t-test’ or a `paired t-test’?
State which of these you would use, and give reasons for your answer. (2 marks)
(c) For the test that you chose in (b), state the statistical assumptions underlying the analysis.
Do these assumptions seem valid in this case? Answer `Yes’ or `No’, and give reasons for
your answer. You must show any computer output that you use. (3 marks)
(d) Irrespective of your answer to (c), carry out the test of H0 vs H1, showing all steps in the
decision process (except that you do not need to restate H0 and H1). (5 marks)
Marking Criteria
Marks 0 1 2 3
5(a) Neither hypothesis cor-
rect, or symbols used
but not explained
One hypothesis correct
and symbols explained
Both hypotheses cor-
rect and symbols
explained
5(b) Test to be used incor-
rect, or not speci
ed
Correct test speci
ed,
butjusti
cationinade-
quate
Correct test speci
ed,
andjusti
cationcor-
rect and clear
5(c) Assumptions not
stated, or incorrect
One assumption cor-
rect, but not tested for
validity
All assumptions cor-
rect, but not all tested
for validity
All assumptions cor-
rect, and all correctly
tested for validity
Marks 0 1
5(d) Test statistic and
samplingdistribu-
tion
Statistic and distribution either
missing or incorrect
Statistic and distribution both
correct
Rejection criterion Criterion missing or incorrect Criterion correctly stated
Calculations (soft-
ware or manually)
Calculations wrong or only par-
tially correct
Calculations correct
Decision Decision missing or not justi
ed Decision correctly stated and
justi
ed
Conclusion Missing, or not in context of prob-
lem
Statement in English in context of
problem
7
STA201/401 201530 Assignment 2
Question 6 | STA401 only (18 marks in total)
This question is ONLY for STA401 students. STA201 students do NOT need to submit this
question.
(This is based on a question from Daniel, W.W. and Cross, C.L. (2013). Biostatistics: A
Foundation for Analysis in the Health Sciences. (10th edn) Hoboken, NJ: Wiley.)
Two independent samples of Year 7 students were selected. Sample A was drawn from children
who were caries-free, while Sample B was drawn from children who had lots of caries. The
saliva pH levels were measured for all the children. The data are stored in several forms in the
le saliva.csv in the folder in Interact2 where this Assignment is located. It is suggested that
you look at the
le so that you know the format of the data before you import it to the R
Commander.
(a) You are to use the independent-samples pooled t-test to test whether the mean saliva pH
level of caries-free children di
ers from the corresponding mean for children with lots of
caries. Use a 5% level of signi
cance in all your testing.
i. State the statistical assumptions about the data that underly the independent-samples
pooled t-test. (2 marks)
ii. Are the assumptions valid for the supplied data? (You may use the R Commander
to help you decide this.) Justify your answer. (8 marks)
iii. Irrespective of your answer to ii., perform the independent-samples pooled t-test to
answer the research question at the beginning of section (a). You should follow
the standard steps of hypothesis testing (although you are encouraged to use the R
Commander to do the calculations). Your answer must include a sentence in English
that is useful to the researchers (who do not know any Statistics). (6 marks)
(b) Suppose that you need to
nd a 95% con
dence interval for the variance of a population,
and the sample of 20 observations that you have available to you clearly indicates that
the data are not Normally distributed. Brie
y suggest a technique that you might use to
calculate the 95% con
dence interval. [Note that you need only to suggest the technique
and include a couple of lines of justi
cation. You do not need to perform any calculations.]
(2 marks)
Marking Criteria
Marks 0 1 2 3
6(a)i. All assumptions incor-
rectly stated
One assumption cor-
rectly stated
All assumptions cor-
rectly stated
6(a)ii. (See below)
6(a)iii. (See below)
6(b) No appropriate tech-
nique suggested
An appropriate technique suggested, but
justification incorrect
or missing
An appropriate technique suggested, and
correctlyjustified
8
STA201/401 201530 Assignment 2
Marks 0 1
6(a)ii. First assumption No or incorrect justification for answer
Answer and justification correct
Second assumption No or incorrect justification
Question 1
Cans of Fizzy Cola are believed to have contents that are Normally distributed with a mean of 305 ml and a standard deviation of 10 ml.
(a) If a random sample of 25 cans is selected from a day’s production, what is the probability that the 12th can contains more than 300 ml?
Here, we have to calculate p(X>300)
Here, we have to use the z-score formula given as below;
Z = ( 300 – 305) / 10 = -5/10 = -0.5
P(z>-0.5) = 0.691462
Required probability = 0.691462
(b) If the average of the contents of the 25 cans is calculated, what is
- The standard error of this average?
Standard error = stdev. / sqrt(n) = 10 / sqrt(25) = 2
- The probability that this average is greater than 300 ml?
Solution:
Z = (X – mean ) / [ stdev./sqrt(n)
Z = (300 – 305) / [ 10 / sqrt(25)]
Z = -2.5
P(x>300) = P (z>-2.5) = 0.99379
Required probability = 0.99379
(c) What is the probability that at least 20 of the 25 cans have contents greater than 300 ml?
Z = (X – mean ) / [ stdev./sqrt(n)
Z = (300 – 305) / [ 10 / sqrt(20)]
Z = -2.236
P(x>300) = P (z>-2.236) = 0.9873
Required probability = 0.9873
Question 2
Suppose that, in 10 years of use throughout Australia, one particular animal vaccine has been successful in treating an animal disease in 66% of cases (i.e., on average, 66% of animals have recovered), and this percentage has become accepted as the industry ‘standard’.
(a) JaneyWerakso has 64 animals suffering from this disease. If she administers the vaccine in the approved manner, what is the standard error of the proportion of animals that will recover?
Standard error of proportion = sqrt(pq/n) = sqrt( 0.66*0.34/64) = 0.003506
(b) After treatment, Janey finds that 39 of her animals recover from the disease. What proportion of Janey’s animals recover from this disease?
Required proportion = 39/64 = 0.6094
(c) What is the probability of getting this proportion (or less) of recoveries from 64 animals if the vaccine really does have a 66% recovery rate?
Solution:
Z = ( 64*0.6094 – 64*0.66) / sqrt(64*0.66*0.34) = -0.8545
P(z< -0.8545) = 0.1964
Required probability = 0.1964
(d) If p represents the true probability that an animal recovers from the disease when the vaccine is administered, use the data from Janey’s experiment to test H0: p = 0.66 vs H1: p 6= 0.66 at the 5% level of significance. (5 marks) (e) Janey is pretty annoyed that less than 66% of her animals recovered, and wants to know why we didn’t test H0: p = 0.66 vs H1: p < 0.66 because, after all, less than 66% recovered. Is this a valid thing to do? Answer ‘yes’ or ‘no’, and justify your answer.
Number of items of interest = 64*0.6094 = 39
Z Test of Hypothesis for the Proportion | |
Data |
|
Null Hypothesis p = |
0.66 |
Level of Significance |
0.05 |
Number of Items of Interest |
39 |
Sample Size |
64 |
Intermediate Calculations |
|
Sample Proportion |
0.609375 |
Standard Error |
0.0592 |
Z Test Statistic |
-0.8550 |
Two-Tail Test |
|
Lower Critical Value |
-1.9600 |
Upper Critical Value |
1.9600 |
p-Value |
0.3926 |
Do not reject the null hypothesis |
(e) In the light of Janey’s concerns, would it be valid to take another large sample of animals and test H0: p = 0.66 vs H1: p < 0.66? Justify your answer.
Sample Size Determination | |
Data |
|
Estimate of True Proportion |
0.66 |
Sampling Error |
0.0592 |
Confidence Level |
95% |
Intermediate Calculations |
|
Z Value |
-1.9600 |
Calculated Sample Size |
245.9663 |
Result |
|
Sample Size Needed |
246 |
Yes, it would be valid to take another large sample size of 246 animals.
Question 3
Suppose that a reputable survey agency is engaged to estimate the proportion of Australian electors who believe that, to reduce carbon emissions, a nuclear power plant should be built somewhere in Australia.
(a) Under the assumption that everyone who is contacted gives a reply of ‘In favour’ or ‘Opposed/Don’t Know’, how large a sample must be taken if the estimate obtained from the survey is to have a margin of error of at most 2%? Use a 95% confidence interval in calculating the margin of error, and assume that absolutely nothing is known about the proportion of Australians in favour.
We absolutely nothing is known about the proportion of Australians in Favour, in this case, we take p = 0.5 or 50%
Sample Size Determination | |
Data |
|
Estimate of True Proportion |
0.5 |
Sampling Error |
0.02 |
Confidence Level |
95% |
Intermediate Calculations |
|
Z Value |
-1.9600 |
Calculated Sample Size |
2400.9118 |
Result |
|
Sample Size Needed |
2401 |
(b) Repeat (a) if it is believed that the proportion of Australian electors in favour is not more than 30%.
Sample Size Determination | |
Data |
|
Estimate of True Proportion |
0.3 |
Sampling Error |
0.02 |
Confidence Level |
95% |
Intermediate Calculations |
|
Z Value |
-1.9600 |
Calculated Sample Size |
2016.7659 |
Result |
|
Sample Size Needed |
2017 |
(c) Repeat (a) if it is thought that 20% of people contacted will refuse to answer, thereby providing no information. (Make the assumption that these people think like the other 80%, but just don’t want to answer.)
Sample Size Determination | |
Data |
|
Estimate of True Proportion |
0.2 |
Sampling Error |
0.02 |
Confidence Level |
95% |
Intermediate Calculations |
|
Z Value |
-1.9600 |
Calculated Sample Size |
1536.5835 |
Result |
|
Sample Size Needed |
1537 |
(d) Suppose that a random sample of 1000 Australian electors is asked whether they believe that, to reduce carbon emissions, a nuclear power plant should be built somewhere in Australia. After eliminating the 221 people who refuse to answer, 275 of the remainder say that they support building a nuclear power plant. Calculate a 95% confidence interval for the true proportion, p, of Australian electors who believe that, to reduce carbon emissions, a nuclear power plant should be built somewhere in Australia.
Confidence Interval Estimate for the Proportion | |
Data |
|
Sample Size |
1000 |
Number of Successes |
275 |
Confidence Level |
95% |
Intermediate Calculations |
|
Sample Proportion |
0.275 |
Z Value |
-1.9600 |
Standard Error of the Proportion |
0.0141 |
Interval Half Width |
0.0277 |
Confidence Interval |
|
Interval Lower Limit |
0.2473 |
Interval Upper Limit |
0.3027 |
(e) Use the confidence interval obtained in (d) to test H0: p = 0.30 vs H1: p 6= 0.30 at the 5% level of significance. You do not have to carry out the formal steps of hypothesis testing, but your answer must say whether or not you reject H0, and give reasons for your answer.
Solution:
Here, we do not reject the null hypothesis, because p = 0.30 is lies between the confidence interval (0.2473, 0.3027).
Question 4
A researcher is investigating the effectiveness of a new diet that is being considered for raising chickens. He wants to measure the weight gained by chicks on this diet in the 14 days between the 7th and 21st days of life. On a given date, a random sample of 25 7-days-old chicks is selected, and their weights are measured then and again on the 21st day. The weight gains (in kg) are recorded in the file chicken weights.csv in the folder in Interact2 where this Assignment is located. The researcher wants to estimate the mean weight gained (in kg), µ, by chicks on this diet. He asks you to calculate a 95% confidence interval for the value of µ, and tells you to use the t-distribution.
(a) What assumptions are required to be met for the use of the t-distribution? Do they seem valid here? Justify your answer.
If we do not given the population standard deviation, then we use t distribution and if we are given the population standard deviation, then we use the z distribution. Here, we do not given the information about the population standard deviation, so we need to use the t distribution.
(b) Irrespective of your answer to (a), calculate a 95% confidence interval for µ.
Confidence Interval Estimate for the Mean | |
Data |
|
Sample Standard Deviation |
0.202607996 |
Sample Mean |
0.766 |
Sample Size |
25 |
Confidence Level |
95% |
Intermediate Calculations |
|
Standard Error of the Mean |
0.040521599 |
Degrees of Freedom |
24 |
t Value |
2.0639 |
Interval Half Width |
0.0836 |
Confidence Interval |
|
Interval Lower Limit |
0.68 |
Interval Upper Limit |
0.85 |
(c) The mean weight gain over 14 days under a standard diet is 0.85 kg. (This figure has been established over several years, and can be regarded as a known constant.) Test at the 1% significance level whether the new diet has a mean weight gain that is greater than 0.85 kg. Give all the steps of the hypothesis testing process.
t Test for Hypothesis of the Mean | |
Data |
|
Null Hypothesis m= |
0.85 |
Level of Significance |
0.01 |
Sample Size |
25 |
Sample Mean |
0.766 |
Sample Standard Deviation |
0.202607996 |
Intermediate Calculations |
|
Standard Error of the Mean |
0.0405 |
Degrees of Freedom |
24 |
t Test Statistic |
-2.0730 |
Upper-Tail Test |
|
Upper Critical Value |
2.4922 |
p-Value |
0.9755 |
Do not reject the null hypothesis |
Question 5
Biostatistics: A Foundation for Analysis in the Health Sciences. (10th edn) Hoboken, NJ: Wiley.) Researchers investigated the effect of broadband ultraviolet B (UVB) therapy and a topical cream used together on areas of psoriasis (a skin complaint). One of the outcome variables was the Psoriasis Area and Severity Index (PASI). The Table that appears below contains the PASI scores for 20 patients who were assessed at baseline and after eight treatments.
Subject 1 2 3 4 5 6 7 8 9 10 Baseline 5.9 7.6 12.8 16.5 6.1 14.4 6.6 5.4 9.6 11.6 After 8 Treatments 5.2 12.2 4.6 4.0 0.4 3.8 1.2 3.1 3.5 4.9 Subject 11 12 13 14 15 16 17 18 19 20 Baseline 11.1 15.6 6.9 15.2 21.0 5.9 10.0 12.2 20.2 6.2 After 8 Treatments 11.1 8.4 5.8 5.0 6.4 0.0 2.7 5.1 4.8 4.2
The original question asked ‘Do these data provide evidence, at the .01 level of significance, to indicate that the combination therapy reduces PASI scores?’
(a) State the null and alternative hypotheses that you would test in order to answer this question. If you use symbols other than H0 and H1 (and ‘=’, etc) in your statement, you must explain what those symbols represent.
Null hypothesis: H0: mean for baseline = mean for after 8 treatments
Alternative hypothesis: Ha: mean for baseline ≠ mean for after 8 treatments
(b) If the data satisfy the appropriate statistical assumptions, then these hypotheses can be tested by using a t-test. Would you use a ‘Welch Two Sample t-test’ or a ‘paired t-test’? State which of these you would use, and give reasons for your answer.
We will use the paired t test because we are given a pairs for baseline and after 8 treatments.
(c) For the test that you chose in (b), state the statistical assumptions underlying the analysis. Do these assumptions seem valid in this case? Answer ‘Yes’ or ‘No’, and give reasons for your answer. You must show any computer output that you use.
Here, we have to use the paired sample t test. For this test, we need the data for after and before type. Here, we already have data for after and before. The differences between pairs would be normally distributed.
(d) Irrespective of your answer to (c), carry out the test of H0 vs H1, showing all steps in the decision process (except that you do not need to restate H0 and H1).
Solution:
Paired t Test | |
Data |
|
Hypothesized Mean Difference |
0 |
Level of significance |
0.05 |
Intermediate Calculations |
|
Sample Size |
20 |
DBar |
6.2200 |
Degrees of Freedom |
19 |
SD |
5.0403 |
Standard Error |
1.1271 |
t Test Statistic |
5.5188 |
Two-Tail Test |
|
Lower Critical Value |
-2.0930 |
Upper Critical Value |
2.0930 |
p-Value |
0.0000 |
Reject the null hypothesis |
Question 6
This question is ONLY for STA401 students. STA201 students do NOT need to submit this question. (This is based on a question from Daniel, W.W. and Cross, C.L. (2013). Biostatistics: A Foundation for Analysis in the Health Sciences. (10th edn) Hoboken, NJ: Wiley.) Two independent samples of Year 7 students were selected. Sample A was drawn from children who were caries-free, while Sample B was drawn from children who had lots of caries. The saliva pH levels were measured for all the children. The data are stored in several forms in the file saliva.csv in the folder in Interact2 where this Assignment is located.
(a) You are to use the independent-samples pooled t-test to test whether the mean saliva pH level of caries-free children differs from the corresponding mean for children with lots of caries. Use a 5% level of significance in all your testing.
Solution:
Pooled-Variance t Test for the Difference Between Two Means | |
(assumes equal population variances) | |
Data |
|
Hypothesized Difference |
0 |
Level of Significance |
0.05 |
Population 1 Sample |
|
Sample Size |
15 |
Sample Mean |
7.519333333 |
Sample Standard Deviation |
0.29884221 |
Population 2 Sample |
|
Sample Size |
12 |
Sample Mean |
7.271666667 |
Sample Standard Deviation |
0.20112562 |
Intermediate Calculations |
|
Population 1 Sample Degrees of Freedom |
14 |
Population 2 Sample Degrees of Freedom |
11 |
Total Degrees of Freedom |
25 |
Pooled Variance |
0.0678 |
Standard Error |
0.1009 |
Difference in Sample Means |
0.2477 |
t Test Statistic |
2.4557 |
Two-Tail Test |
|
Lower Critical Value |
-2.0595 |
Upper Critical Value |
2.0595 |
p-Value |
0.0214 |
Reject the null hypothesis |