## Question

## Transcribed Text

1. How well materials conduct heat matters when designing houses, for example. Conductivity is measured in terms of watts of heat power transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material. In these units, glass has a conductivity of about 1. The National Institute of Standards and Technology (NIST) provides data on properties of materials. Here is a simple random sample of 11 NIST measurements of the heat conductivity of a particular type of glass:
1.11 1.07 1.11 1.07 1.12 1.08 1.08 1.15 1.18 1.18 1.12
Sample mean = 1.1155
Sample standard deviation = 0.0403
a. Using the probability plot and a boxplotIs it reasonable to use the t procedures? Why or why not?
Distribution of Heat Conductivity
1.06 1.08 1.10 1.12 1.14 1.16 1.18
Heat Conductivity (in Watts per meter squared per Celcius)
If the answer is no, do not do parts b, c, and d.
b. If your conclusion in part (a) is yes, give a 95% confidence interval for the mean conductivity. (Make sure your interpretation of the 95% confidence interval is in a complete sentence in the context of the problem.)
c. If your conclusion in part (a) is yes, give a 90% confidence interval for the mean conductivity. (Make sure your interpretation of the 90% confidence interval is in a complete sentence in the context of the problem.)
d. If your conclusion in part (a) is yes, do the data give convincing evidence that the mean heat conductivity of this particular type of glass is less than 1.15 at a 0.01 significance level?
i. State the null and alternative hypotheses.
ii. State the significance level for this problem.
iii. State the test statistic.
iv. State the critical value.
v. State whether you reject or do not reject the null hypothesis.
vi. State your conclusion in context of the problem.
vii. If the true mean was 1.15, did you make an error? If so, which error? (Answer in complete sentences. You need to explain why or why not you made an error as well as identifying the type of error (Type I or II) if you did make an error. DO NOT answer these questions with a simple "Yes" or "No"; if you do, it will be marked as incorrect.)
99
95
90
80 70
60 50 40 30 20
10 5
1
1.00 1.05
Normal - 95% CI
1.10 1.15 1.20 1.25
Probability Plot of Heat Conductivity
Heat Conductivity ( in Watts per square meter per Celsius)
Mean 1.115 StDev 0.04034 N 11 A D 0.460 P-Value 0.209
Percent
2. Here’s a new idea for treating advanced melanoma, the most serious kind of skin cancer. Genetically engineer white blood cells to better recognize and destroy cancer cells, then infuse these cells into patients.
An outcome in the cancer experiment described above is measured by a test for the presence of cells that trigger an immune response in the body and so may help fight cancer. Here is a simple random sample of 11 subjects: counts of active cells per 100,000 cells before and after infusion of the modified cells. The difference (after minus before) is the response variable.
Before
After Difference(After-Before)
14 0 1 0 0 0 0 20 1 6 0 41 7 1 215 20 700 13 530 35 92 108 27 7 0 215 20 700 13 510 34 86 108
Sample mean = 156.4
Sample standard deivation = 234.3
a. Using the probability plot and a boxplot of the differences, is it reasonable to use the t procedures? Why or why not?
Distribution of Difference = (After-Before)
b, and c
0 100 200 300 400 500 600 700
Difference = (After-Before) (in counts of active cells/100,000 cells)
b. If your conclusion in part (a) is yes, do a 99% confidence interval about the mean difference of active cells after treatment and before treatment (the 99% confidence interval on the difference (after- before). (Make sure your interpretation of the 99% confidence interval is in a complete sentence in the context of the problem.)
c. If your conclusion in part (a) is yes, do the data give convincing evidence that the count of active cells is higher after treatment at a 0.05 significance level.
i. State the null and alternative hypotheses.
ii. State the significance level for this problem.
iii. State the test statistic.
iv. State the critical value.
v. State whether you reject or do not reject the null hypothesis.
vi. State your conclusion in context of the problem.
vii. If the true mean was not 0, did you make an error? If so, which error? (Answer in complete sentences. You need to explain why or why not you made an error as well as identifying the type of error (Type I or II) if you did make an error. DO NOT answer these questions with a simple "Yes" or "No"; if you do, it will be marked as incorrect.)
99
95
90
80 70
60 50 40 30 20
10 5
1
-500
Normal - 95% CI
If the answer is no, do not do parts
0 500 1000
Probability Plot of Difference = (After-Before)
Difference = (After-Before) (in counts of active active cells/100,000 cells)
Mean 156.4 StDev 234.3 N 11 A D 1.455 P-Value <0.005
Percent
3. A physical therapist wanted to know whether the mean step pulse of men (m) was less than the mean step pulse of women (w). She randomly selected 51 men and 70 women to participate in the study. Each subject was required to step up and down onto a 6-inch platform for 3 minutes. The pulse of each subject (in beats per minute) was then recorded. After the data were entered into MINITAB, the following results were obtained. (Use the conservative df.
Two Sample T-Test and Confidence Interval
Two Sample T for Men vs Women
N Mean StDev SE Mean
Men 51 112.3 11.3 1.6
women 70 118.3 14.2 1.7
95% CI for mu Men – mu Women: (-10.7, -1.5)
T-Test mu Men = mu Women (vs < 0): T = -2.61 P-Value = 0.0051
a. State the null and alternative hypothesis.
b. State whether you would reject or not reject the null hypothesis at = 0.01 based on the MINITAB output.
c. A 95% confidence interval is given above in the MINITAB output. However, the therapist wanted a 90% confidence interval about the difference of men’s mean step pulse rate and
12 women’s mean step pulse rate. Given the standard error is 2.32 (that is, s2 + s2 = 2.32),
calculate the 90% Confidence Interval using the conservative df
4. Makers of generic drugs must show that they do not differ significantly from the “reference” drugs that they imitate. One aspect in which drugs might differ is their extent of absorption in the blood. Table 17.6 gives data taken from 20 healthy nonsmoking male subjects for one pair of drugs. This is a matched pairs design. Numbers 1 to 20 were assigned at random to the subjects. Subjects 1 to 10 received the generic drug first, and Subjects 11 to 20 received the reference drug first. In all cases, a washout period separated the two drugs so that the first had disappeared from the blood before the subject took the second.
TABLE 17.6 Absorption extent for two versions of a drug
nn
12
You will note that subject 15 has been deleted as the input was incorrect.
Sample Mean = -163
Sample Standard Deviation = 936
a. Use the probability plot and a boxplot of the differences to see if it is reasonable to use the t procedures?
Distribution of Differences in Blood Absorption
between Reference and Generic Drug
-2000 -1000 0 1000 2000
Difference in Blood Absorptions between Reference and Generic Drug
99
95
90
80 70
60 50 40 30 20
10 5
1
Mean -162.8 StDev 935.9 N19
Probability Plot of Difference = (Ref - Gen)
Normal - 95% CI
-3000
-2000
-1000 0
1000 2000
3000
Difference = (Ref - Gen)
A D P-Value
0.233 0.765
If the answer is no, do not do parts b and c.
b. If your conclusion in part (a) is yes, do a 99% confidence interval about the mean difference of absorption between the reference drug and the generic drug. (Make sure your interpretation of the 99% confidence interval is in a complete sentence in the context of the problem.)
c. If your conclusion in part (a) is yes, do the data give convincing evidence that the drugs differ in absorption at a 0.05 significance level.
i. State the null and alternative hypotheses.
ii. State the significance level for this problem.
iii. State the test statistic.
iv. State the critical value(s).
v. State whether you reject or do not reject the null hypothesis.
vi. State your conclusion in context of the problem.
vii. If the true mean was 0, did you make an error? If so, which error? (Answer in complete sentences. You need to explain why or why not you made an error as well as identifying the type of error (Type I or II) if you did make an error. DO NOT answer these questions with a simple "Yes" or "No"; if you do, it will be marked as incorrect.)
Percent

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Question 1:

a. Using the probability plot and a box plots it reasonable to use the t procedures? Why or why not?

Using the below two plots we can say it is reasonable to use the t procedures because the normal probability plots shows that points close to the straight line and inside the 95% confidence limits. Also the standard deviation is very small in this case, It means data is normally distributed. The p-value associated with normal probability plot is 0.209 which is bigger than 0.05 so we can conclude that the data for heat conductivity is normally distributed. The same can also be seen from the box plot where we can see that the data is equally spread without any outliers. These both plots show that the assumption of using the t procedures are met and we can use the t distribution on this data.

b. If your conclusion in part (a) is yes, give a 95% confidence interval for the mean conductivity. (Make sure your interpretation of the 95% confidence interval is in a complete sentence in the context of the problem.)

95% confidence interval for mean conductivity is given below:

Lower Limit = Sample Mean – Critical Value of t at n-1 degree of freedom*Standard Error

Upper Limit = Sample Mean + Critical Value of t at n-1 degree of freedom*Standard Error

Here we have,

Sample mean = 1.1155

Sample standard deviation = 0.04034 and n = 11

Standard Error = Sample standard deviation/sqrt(n) = 0.04034/sqrt(11) =0.012162968

Critical value at 5% level of significance with n-1=11-1=10 degree of freedom is ±2.22813

Lower Limit = 1.1155 - 2.22813*0.012162968 = 1.088399

Upper Limit = 1.1155 + 2.22813*0.012162968 = 1.142601

We are 95% confident that population mean heat conductivity of a glass lies between 1.088399 and 1.142601....

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