An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection, and the second group received the drug. After a 30-day period, the proportions of survivors, p̂1 and p̂2, in the two groups were found to be 0.4 and 0.60, respectively.
Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.
z = ?? (please do not use (p1-p2) / SE as it does not work. I have tried this multiple times)
z < = -1.65
z > = NONE
(b) Use a 95% confidence interval to estimate the actual difference (p1 − p2) in the survival rates for the treated versus the control groups. (Round your answers to two decimal places.)
In: Statistics and Probability
In a single slit experiment, the central maximum is:
|
The same size as all of the other maximums. |
|
|
Twice the size of the other maximums. |
|
|
Half the size of the other maximums. |
|
|
Just slightly larger than the other maximums. |
Electrons would be ejected from silver if the incident light had a frequency of:
|
1.03 x 1015 Hz |
|
|
1.02 x 1015 Hz |
|
|
1.01 x 1015 Hz |
|
|
None of the options is correct |
What is the mass equivalent of an x-ray with a frequency of 5.63 x 1017 Hz?
|
4.14 x 10-33 kg |
|
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2.59 x 10-14 kg |
|
|
1.24 x 10-24 kg |
|
|
None of the options is correct. |
In: Physics
An experiment was conducted to determine the effect of a high salt mean on the systolic blood pressure (SBP) of subjects. Blood pressure was determined in 12 subjects before and after ingestion of a test meal containing 10.0 gms of salt. The data obtained were:
|
Subject |
SBP before meal |
SBP after meal |
|
1 |
120 |
147 |
|
2 |
130 |
140 |
|
3 |
139 |
148 |
|
4 |
120 |
115 |
|
5 |
123 |
122 |
|
6 |
140 |
157 |
|
7 |
131 |
144 |
|
8 |
123 |
134 |
|
9 |
125 |
140 |
|
10 |
130 |
165 |
|
11 |
131 |
133 |
|
12 |
142 |
153 |
In: Statistics and Probability
In this week's experiment, the heat of vaporization of liquid
nitrogen is determined by measuring the change in temperature of a
known sample of warm water when liquid nitrogen is added.
In one experiment, the mass of water is 104 grams,
the initial temperature of the water is
69.3oC, the mass of liquid nitrogen
added to the water is 60.6 grams, and the final
temperature of the water, after the liquid nitrogen has vaporized,
is 41.3oC.
Specific heat of water = 4.184 J K-1g-1
How much heat is lost by the warm water?
Heat lost = J
What is the heat of vaporization of nitrogen in J
g-1?
Heat of vaporization
= J
g-1
What is the molar heat of vaporization of nitrogen?
Molar heat of vaporization
=
J mol-1
Trouton's constant is the ratio of the enthalpy (heat) of
vaporization of a substance to its boiling point (in K). The
constant is actually equal to the entropy change for the
vaporization process and is most often a measure of the entropy in
the liquid state. The value of the constant usually lies within the
range 70 to 90 J K-1mol-1, with a value
toward the lower end indicating high entropy in the liquid
state.
The normal boiling point of liquid nitrogen is -196oC.
Based upon your results above, what is the value of Trouton's
constant?
Trouton's constant
= J
K-1mol-1
In: Chemistry
Consider the experiment of rolling two dice and the following events:
A: 'The sum of the dice is 8' and B: 'The first die is an odd number' and C: "The difference (absolute value) of the dice is 2"
Find (a) p(A and B) (HINT: You cannot assume these are independent events.)
(b) p(A or B)
(c) Are A and B mutually exclusive events? Explain.
(d) Are A and B independent events? Explain.
(e) Are B and C independent events? Explain.
In: Statistics and Probability
. A new vaccination is being used in a laboratory experiment to investigate whether it is effective. There are 252 subjects in the study. Is there sufficient evidence to determine if vaccination and disease status are related?
Vaccination Status Diseased Not Diseased Total
Vaccinated 51) 54) 105)
Not Vaccinated 54) 73) 147)
Total (125) (127) ( 252)
State the null and alternative hypothesis.
Find the value of the test statistic. Round your answer to three decimal places.
Find the degrees of freedom associated with the test statistic for this problem.
Find the critical value of the test at the 0.01 level of significance. Round your answer to three decimal places.
Make the decision to reject or fail to reject the null hypothesis at the 0.01 level of significance.
State the conclusion of the hypothesis test at the 0.01 level of significance.
In: Statistics and Probability
Here is the data for our experiment.
The data are the SMUT scores of the students in each group. Notice that we have a different number (n) for the lecture group. This is to show you that we can have uneven sets of data for ANOVA. Note: If we were doing a real study, we would have larger n’s. Enter the data into the Excel spread sheet, SPSS or your calculator
| Independent Variable | Data |
| Computer (n1= 8) | 85, 87, 91, 78, 90, 83, 86, 89 |
| Lecture (n2= 7) | 79, 76, 75, 79, 75, 72, 81, |
| Cooperative (n3= 8) | 80, 99, 87, 84, 83, 77, 76, 80 |
This assignment is part of my ANOVA Exercise, I will please need help in completing it.
Thanks
In: Statistics and Probability
The authors of a paper describe an experiment to evaluate the effect of using a cell phone on reaction time. Subjects were asked to perform a simulated driving task while talking on a cell phone. While performing this task, occasional red and green lights flashed on the computer screen. If a green light flashed, subjects were to continue driving, but if a red light flashed, subjects were to brake as quickly as possible. The reaction time (in msec) was recorded. The following summary statistics are based on a graph that appeared in the paper. n = 61 x = 530 s = 75 (a) Assuming that this sample is random/representative of the population, what other assumptions need to be true before we can create a confidence interval? Yes, because the population distribution is normal. No, because n < 30 No, because either np̂ < 10 or n(1−p̂) < 10 Yes, because np̂ ≥ 10 and n(1−p̂)≥ 10 Yes, because n ≥ 30 No, because the population distribution is not normal. Changed: Your submitted answer was incorrect. Your current answer has not been submitted. (b) Construct a 98% confidence interval for μ, the mean time to react to a red light while talking on a cell phone. (Round your answers to three decimal places.) , (c) Interpret a 98% confidence interval for μ, the mean time to react to a red light while talking on a cell phone. We are % confident that the mean time to react to a is between and milliseconds. (d) Suppose that the researchers wanted to estimate the mean reaction time to within 5 msec with 95% confidence. Using the sample standard deviation from the study described as a preliminary estimate of the standard deviation of reaction times, compute the required sample size. (Round your answer up to the nearest whole number.) n = You may need to use the appropriate table in Appendix A to answer this question.
In: Statistics and Probability
Hello,
In an experiment to determine the concentration of glucose in a sample, we are supposed to make a glucose assay with glucose solutions of known concentration. The absorbance for each solution of known concentration will be plotted, so that we can use the line of absorbance rates to later find the concentration of our unknown sample. We need to make our own dilutions for the assay.
We are given a stock glucose solution of 10mg/mL (1000mg/dL). My group wants to make solutions with concentrations 0mg/dL; 100mg/dL; 200mg/dL; 300mg/dL; 400mg/dL; and 500mg/dL. We are unsure of the best way to make the dilutions.
Would we make a 100mg/dL dilution be adding 1 part of stock and 9 parts deionized water? And a 200mg/dL dilution by adding 2 part stock and 8 parts deionized water; 300mg/dL as 3 parts stock and 7 parts deionized water, and so on? I'm unsure of how to do this.
Thank you!
In: Chemistry
An experiment was carried out to investigate the effect of species (factor A, with I = 4) and grade (factor B, with J = 3) on breaking strength of wood specimens. One observation was made for each species—grade combination—resulting in SSA = 444.0, SSB = 424.6, and SSE = 122.4. Assume that an additive model is appropriate. (a) Test H0: α1 = α2 = α3 = α4 = 0 (no differences in true average strength due to species) versus Ha: at least one αi ≠ 0 using a level 0.05 test. Calculate the test statistic. (Round your answer to two decimal places.) f = 1 What can be said about the P-value for the test? P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 P-value < 0.001 State the conclusion in the problem context. Reject H0. The data suggests that true average strength of at least one of the species is different from the others. Fail to reject H0. The data does not suggest any difference in the true average strength due to species. Reject H0. The data does not suggest any difference in the true average strength due to species. Fail to reject H0. The data suggests that true average strength of at least one of the species is different from the others. (b) Test H0: β1 = β2 = β3 = 0 (no differences in true average strength due to grade) versus Ha: at least one βj ≠ 0 using a level 0.05 test. Calculate the test statistic. (Round your answer to two decimal places.) f = 4 What can be said about the P-value for the test? P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 P-value < 0.001 State the conclusion in the problem context. Reject H0. The data does not suggest any difference in the true average strength due to grade. Fail to reject H0. The data does not suggest any difference in the true average strength due to grade. Reject H0. The data suggests that true average strength of at least one of the grades is different from the others. Fail to reject H0. The data suggests that true average strength of at least one of the grades is different from the others.
In: Statistics and Probability