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The average test score for the statistics course is 81%. A professor wants to see if...

The average test score for the statistics course is 81%. A professor wants to see if the average test score for students who are given colored pens on the first day of class is different. The test scores for the 12 randomly selected students who were given the colored pens are shown below. Assume that the distribution of the population is normal.

82, 59, 64, 78, 66, 73, 79, 61, 65, 54, 85, 58

What can be concluded at the the αα = 0.10 level of significance level of significance?

  1. For this study, we should use? (t-test for a population mean or z-test for a population proportion)
  2. The null and alternative hypotheses would be:

H0: (p or μ) (< = > ≠)

H1: (μ or p) ( < > ≠ =)

  1. The test statistic? t or z  = _____ (please show your answer to 3 decimal places.)
  2. The p-value = ______ (Please show your answer to 4 decimal places.)
  3. The p-value is? (> or ≤) α
  4. Based on this, we should? (fail to reject, accept, reject) the null hypothesis.
  5. Thus, the final conclusion is that ...
    • The data suggest the populaton mean is significantly different from 81 at αα = 0.10, so there is statistically significant evidence to conclude that the population mean test score for students who are given colored pens at the beginning of class is different from 81.
    • The data suggest the population mean is not significantly different from 81 at αα = 0.10, so there is statistically insignificant evidence to conclude that the population mean test score for students who are given colored pens at the beginning of class is equal to 81.
    • The data suggest that the population mean test score for students who are given colored pens at the beginning of class is not significantly different from 81 at αα = 0.10, so there is statistically insignificant evidence to conclude that the population mean test score for students who are given colored pens at the beginning of class is different from 81.
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Answer #1

Here we have n = 12 sample meanid = Exi 824 п = 68.667 Standard deviation is). ( 7-zJ? 12-1) +158-68.6673) 2 10:3602 (a). For

Test statistics t = x=4 68.667-81 10.3602/12 slon -[-4.124 Pualue Pvalue at tal-4.124) and df n=121 = 1 Idishlt, df, tails) =

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