cant effect using a two-tailed test with a s: 05? c, Comparing your answer for p

Question

cant effect using a two-tailed test with a s: 05? c, Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test? he spotlight effect refers to overestimating the extent which others notice your appearance or behavior, pecially when you commit a social faux pas. Effec tively, you feel as if you are suddenly standing in a potlight with everyone looking. In one demonstration of this phenomenon, Gilovich, Medvec, and Sav itsky (2000) asked college students to put on a Barry Manilow T-shirt that fellow students had previously judged to be embarrassing. The participants were then led into a room in which other students were already participating in an experiment. After a few minutes the participant was led back out of the room and was allowed to remove the shirt. Later, each participant was asked to estimate how many people in the room had noticed the shirt. The individuals who were in the room were also asked whether they noticed the shirt In the study, the participants significantly overesti- mated the actual number of people who had noticed a. In a similar study using a sample of n 9 partici- pants, the individuals who wore the shirt produced an average estimate of M 6.4 with SS 162 The average number who said they noticed was 8.1. Is the estimate from the participants signifi cantly different from the actual number? Test the null hypothesis that the true mean is 3.1 using a two-tailed test with a .05. b. Is the estimate from the participants significantly higher than the actual number (u 3.1)? Use a one-tailed test with 14. Many animals, including humans, tend to avoid direct eye contact and even patterns that look like eyes Some insects, including moth have evolved eye-spot wings to help ward off predators patterns on the Scaife (1976) reports a study examining how eye-spot

Explanation / Answer

here std deivation =(SS/(n-1))1/2 =4.5

hence std error of mean =std deviation/(n)1/2 =1.5

therefore test stat t=(X-mean)/std error =(6.4-3.1)/1.5=2.2

for 8 degree of freedom for two tailed test critical calue of t=+/-2.306

as our test stat does not fall into critical refion we can not reject null hypothesis.

b) for one tailed test for 8 degree of freedom for two tailed test critical calue of t =1.8595

as  our test stat does fall into critical refion we reject null hypothesis

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