An educational psychologist hypothesized that presenting learning material in a

Question

An educational psychologist hypothesized that presenting learning material in a hypertext form would result in better learning than traditional text. Hypertext is electronic text containing key words that are hyperlinks, so that when they are clicked the reader moves to a portion of the text describing the key word. In this way, a reader can move about the text at will studying various topics by clicking on key words. In contrast, traditional text requires a student to read through the material in a fixed, linear fashion. The psychologist also wanted to compare hyperlinks that were structured after an experts knowledge of the domain with randomly structured hyperlinks.

He randomly assigned 45 astronomy students to three learning conditions: Expert Hypertext, Random Hypertext, and Normal Text, which he designated as as A1, A2, and A3, where they read and studied material on star formation. The dependent variable was scores on a test over the studied material.

The spreadsheet shows the results of the study. The psychologist performed an ANOVA on the data to test his hypothesis. Notice that the test scores are given in columns B-D and rows 2-16 for the three treatment groups. Look over the summary statistics in the various cells of the table. The means are also displayed in a bar graph chart to the right.

1a) What is the null hypothesis?

A)  A1=A2

B)  A1A2A3

C) A1=A2=A3

D)  A2=A3

1b) What was the conclusion of the researcher?

A) Retain Ho

B) Reject Ha

C) Reject Ho

D) Retain Ha

Subject A1 A2 A3 SSW1 SSW2 SSW3 SSBG Y SSTotal 1 85 55 55 6.760 227.004 489.884 34.418 85 71.684 2 87 63 65 21.160 49.938 147.218 41.818 87 109.551 3 92 76 85 92.160 35.204 61.884 0.360 92 239.218 4 78 72 78 19.360 3.738 0.751 78 2.151 5 74 78 76 70.560 62.938 1.284 74 6.418 6 71 58 95 129.960 145.604 319.218 71 30.618 7 78 80 85 19.360 98.671 61.884 78 2.151 8 95 66 78 158.760 16.538 0.751 95 341.018 9 90 75 65 57.760 24.338 147.218 90 181.351 10 85 82 56 6.760 142.404 446.618 85 71.684 11 75 55 88 54.760 227.004 118.084 75 2.351 12 75 85 89 54.760 223.004 140.818 75 2.351 13 88 65 95 31.360 25.671 319.218 88 131.484 14 78 63 82 19.360 49.938 23.684 78 2.151 15 85 78 65 6.760 62.938 147.218 85 71.684 Mean 82.400 70.067 77.133 49.973 92.996 161.716 76.596 55 463.684 Standard Deviation 7.557 10.105 13.208 63 183.151 n 15.000 15.000 15.000 76 0.284 Grand Mean 76.533 304.684 72 20.551 78 2.151 58 343.484 80 12.018 66 110.951 Source df SS MS F p-value 75 2.351 A (between) 2 1148.933 574.467 5.279 0.009 82 29.884 S/A (within) 42 4570.267 108.816 55 463.684 Total 44 5719.200 85 71.684 65 133.018 Anova: Single Factor 63 183.151 78 2.151 SUMMARY 55 463.684 Groups Count Sum Average Variance 65 133.018 A1 15 1236 82.4 53.542857 85 71.684 A2 15 1051 70.0667 99.638095 78 2.151 A3 15 1157 77.1333 173.26667 76 0.284 95 341.018 85 71.684 ANOVA 78 2.151 Source of Variation SS df MS F P-value F crit 65 133.018 Between Groups 1148.9 2 574.467 5.279 0.009 3.220 56 421.618 Within Groups 4570.3 42 108.816 88 131.484 89 155.418 Total 5719.2 44 95 341.018 82 29.884 65 133.018 76.53333 5719.200

Explanation / Answer

1a) Null hypothesis is Ho: muA1=muB2=muC3.

Answer is C

1b). Here pvalue =0.009 is less than alpha=0.05

We Reject Ho

Answer is C

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