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https//edugen.wileyplus.com/edugen/lti/main.uni C Search eWileyPLUS Section 4 WileyPLUS xCould owning a cat as a child..Welcome to Workstation Lock, Statistics: Unlocking the Power of Data, 2e PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES | Chapter 4, Section 3, Exercise 111 Electrícal Stimulation for Fresh Insight? We introduce a study in which 40 participants are trained to solve problems in a certain way and then asked to solve an unfamiliar problem that requires fresh insight. Half of the participants were randomly assigned to receive electrical stimulation of the brain while the other half (control group) received sham stimulation as a placebo. The results are shown in Table 1 Solved Not solved Treatment Sham Electrical 16 12 Table 1: Does electrical brain stimulation bring fresh insight to a problem? Click here to access Statkey (a) State the null and altemative hypotheses. Let p be the proportion of people who can solve the problem while getting electrical stimulation and let p2 be the proportion of people that can solve the problem with no electrical stimulation. P2 Click here to start 8-23 PM O Type here to searchExplanation / Answer
Solution:-
a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1< P2
Alternative hypothesis: P1 > P2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.20
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.04472
z = (p1 - p2) / SE
z = 4.47
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 4.47.
Thus, the P-value = less than 0.0001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.01), we have to reject the null hypothesis.
Reject H0, From the above test we have sufficient evidence in the favor of the claim that the electric stimulation helps.
e) Yes, becuase the result come from an experiment and they are significant.
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