Homework Answers
The null and alternative hypotheses are
Ho : There is no relationship between Treatment and Problem
Solving
Ha : There is no relationship between Treatment and
Problem Solving
Let Level of Significance α = 0.05
We use Chi Square test of independence
Grand Total of frequencies = 40
To find Expected Frequencies
Expected Frequencies = (Row Total * Column Total)/Grand
Total
| Expected Frequencies (E) | Solved | Not Solved |
| Sham | 8 | 12 |
| Electrical | 8 | 12 |
Following table gives the value of (Observed - Expected)² / Expected
| Solved | Not Solved | |
| Sham | 2 | 1.3333 |
| Electrical | 2 | 1.3333 |
Chi Square Value = ∑[(Observed - Expected)² /
Expected]
χ² = 6.6667
Degrees of freedom = df = (number of rows - 1) * (number of columns
- 1)
Degrees of freedom = df = 1
From chi square table or CHISQ.DIST.RT(X, df) function of
Excel
we find the p-value
p-value = CHISQ.DIST.RT( 6.6667, 1) =
0.0098
p-value = 0.0098
Decision
:
0.0098 < 0.05
that is p-value <= α
Hence we REJECT Ho
Conclusion
:
There exists enough statistical evidence at α = 0.05 to show that
there is a relationship between problem solving and
treatment
Thus electrical simulation helped people solve a problem
that requires fresh insight
Normal
Distribution and Chi Square Distribution
The variables in the problem are nominal variables. The
chi-square test is used mainly
when dealing with a nominal variable
Also it is required to test interaction between two variables, which cannot be tested using normal distribution.
Add Answer to:
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how to show a full diagram. Thanks in advance :)
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) Give values for pUnresolvedE, the proportion of people in the
electrical stimulation group to solve the problem, and
pUnresolvedS, the proportion of people in the sham simulation group
to solve the problem. What is the difference in proportions
pUnresolvedE-pUnresolvedS?
Round your answer to two decimal places.
Unresolved Unresolved el Give values for p the proportion of people in the electrical stimulation group to solve the problem, and.the proportion of people in the sham simulation group to solve the problem....
just need help with problem E. I cannot figure it
out
Lock, Statistics: Unlocking the Power of Data, 2e learned to solve problems by one method, we often have dificulty bringing new insight to similar problems. However, electrical si subjects come up with fresh insight. In a recent experiment' conducted at the University of Sydney in A asked to solve an unfamiliar problem that r half (control group) received sham stimulation as a placebo. The participant the problem while 60%...
leyPLUS × VileyPLUSty n Power of Help I GNMENT If we have learned to solve problems by one method, we often have difficulty bringing new insi ght to similar problems. However, electrical stimulation of the brain appears to help subjects come up with fresh insight. In a a, 40 to solve an unfamiliar half (control group) received sham stimulation as a placebo. The participants did not know in which group they were. In the control the other 20% of the...
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photos for each question are all in a row
(1 point) In the following questions, use the normal distribution to find a confidence interval for a difference in proportions pu - P2 given the relevant sample results. Give the best point estimate for p. - P2, the margin of error, and the confidence interval. Assume the results come from random samples. Give your answers to 4 decimal places. 300. Use 1. A 80% interval for pı - P2 given that...
In Data 4.1 on page 258, we examine a study in which mice were randomly assigned to either a normal light/dark cycle or to have a light on around the clock. According to the paper describing that study, "The global increase in the prevalence of obesity and metabolic disorders coincides with the increase of exposure to light at night." The study is examining whether light at night plays a causal role in the obesity epidemic. In Data 4.1, we examine...
The
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Help with this page is greatly appreciated. I dont understand
how to show a full diagram. Thanks in advance :)
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1. Many companies use a incoming shipments of parts, raw materials, and so on. In the electronics industry, component parts are commonly shipped from suppliers in large lots. Inspection of a sample of n components can be viewed as the n trials of a binomial experimem. The outcome for each component tested (trialD will be that the component is classified as good or defective defective components in the lot do not exceed 1 %. Suppose a random sample of fiver...
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