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Question
Waf/servlet/quiz?quiz, action takeQuiz&quiz; probGuid 387892600b00008uctx kerrykennedy-0063&ck; o 1496872583592 6. A comprehensive one-tailed hypothesis test Aa Aa The Mozart effect refers to the idea that listening to Mozart's music may induce a short-term improvement in the performance of certain kinds of mental tasks. This was first documented in 1993 by Frances Rauscher, Gordon shaw, and Katherine Ky, who published a paper in Nature reporting that students who listened to Mozart's music before doing spatial reasoning tasks performed better than students who did the same tasks after listening to a relaxation tape. You are a cognitive psychology postdoctoral student. You are interested in whether loud music with highly amplified distortion, such as death metal, might have the opposite effect of Mozart's music. To test the hypothesis that listening to death metal decreases performance on spatial reasoning tasks, you administer the paper-folding subtest of the Stanford-Binet Intelligence Test to a random sample of 64. The paper-folding and cutting subtest requires the child to fold and cut papers in a way similar to how the examiner did this task. The subtest raw scores can be converted into normalized standard scores with a mean of so and a standard deviation of You administer the paper-folding subtest to the 64 students while they listen to death metal. The average of this sample is 48.50. Let y equal the true population mean for examinees taking the paper-folding subtest of the stanford-Binet Intelligence Test. The null and alternative hypotheses are formulated as: O Ho: M 2 50, H1: M 50 O Ho: 50, 1: H 2 50 H O Ho: u s 50, H1: H 50 O Ho: u 2 50, H1 k 50Explanation / Answer
Solution:-
The sampling distribution of M is approximated by normal distribution with mean 50 and standard deviation 1.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 50
Alternative hypothesis: < 50
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE) and the z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 1
z = (x - ) / SE
zcritical = - 1.645
Critical region is
z < 1.645
z = - 1.5
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Here is the logic of the analysis: Given the alternative hypothesis ( < 110), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.
Interpret results. Since the z value (- 1.5) is greater than the critical z value (- 1.645), we cannot reject the null hypothesis.
Null hypothesis is correct becuase z value (- 1.5) is greater than the critical z value (- 1.645).
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