Question

Question 24 (4 points) DepScore1 Sum of Squares df F 1.313 Mean Square 23.846 18.158 Sig. 2 74 Between Groups Within Groups T
Question 25 (3 points) Given the above findings, report what you found in terms the null hypothesis using an Alpha level of .
0 0
Add a comment Improve this question Transcribed image text
Answer #1

ANSWER

From given the data it is clearly understood that the naull hypothesis is accepted as,

from f value calculated we find it at 1.313. but the tabulated value of it stands to 3.07

so calculated value is less. so we can accept the null hypothesis.

Another explanation by alpha=0.05 value.

logic is as follows if significant values is greater than or equal to alpha value given then we can say the null hypothesis may be accepted with (1-alpha) % of confidence interval.

Add a comment
Know the answer?
Add Answer to:
Question 24 (4 points) DepScore1 Sum of Squares df F 1.313 Mean Square 23.846 18.158 Sig....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • ANOVA Score   Sum of Squares df Mean Square F Sig. Between Groups 652.875 3 217.625 14.404...

    ANOVA Score   Sum of Squares df Mean Square F Sig. Between Groups 652.875 3 217.625 14.404 .000 Within Groups 543.900 36 15.108     Total 1196.775 39  

  • Question 8 ANOVA Score   Sum of Squares df Mean Square F Sig. Between Groups 1746.100 3...

    Question 8 ANOVA Score   Sum of Squares df Mean Square F Sig. Between Groups 1746.100 3 582.033 47.686 .000 Within Groups 439.400 36 12.206     Total 2185.500 39       Which value below represents the effect size (eta squared) for this analysis? η2 =.83 η2 =.79 η2 =.59 η2 =.69

  •             Sum of Squares df Mean Square F Sig. Between Groups 13.114 4 3.279 2.230 .072...

                Sum of Squares df Mean Square F Sig. Between Groups 13.114 4 3.279 2.230 .072 Within Groups 129.359 88 1.470 Total 142.473 92 Is the relationship between the IV & DV significant? How did you come to this conclusion? Explain the relationship between the IV & DV as if you were reporting it in the “Results” section of a paper that you are writing.

  •                         ANALYSIS OF VARIANCE SOURCE       SUM-OF-SQUARES   DF MEAN-SQUARE &nbsp

                            ANALYSIS OF VARIANCE SOURCE       SUM-OF-SQUARES   DF MEAN-SQUARE     F-RATIO       P Value SEX               8.525                            1        8.525                           5.928       0.020 ERROR         47.455                          33        1.438 TUKEY HSD MULTIPLE COMPARISONS. MATRIX OF PAIRWISE COMPARISON PROBABILITIES:                              1           2           3           4             1          1.000             2          0.021       1.000             3          0.359       0.023      1.000             4          0.054       0.001      0.062     1.000 Does this support the alternative hypothesis and what does this mean? Chart the results.

  • Source Between treatments Within treatments Sum of Squares (Ss) df Mean Square (MS) 2 310,050.00 2,650.00 In some ANOVA summary tables you will see, the labels in the first (source) column are Treatm...

    Source Between treatments Within treatments Sum of Squares (Ss) df Mean Square (MS) 2 310,050.00 2,650.00 In some ANOVA summary tables you will see, the labels in the first (source) column are Treatment, Error, and Total Which of the following reasons best explains why the within-treatments sum of squares is sometimes referred to as the "error sum of squares"? O Differences among members of the sample who received the same treatment occur when the researcher O Differences among members of...

  • Given the following ANOVA table, the F statistic is ANOVA Lab value (at arrival) Sum of...

    Given the following ANOVA table, the F statistic is ANOVA Lab value (at arrival) Sum of Squares Between Groups Within Groups Total 1.939 106.776 2670.214 2776.990 df 2 97 99 Mean Square 5 3.388 27.528 Select one: O a. 27.528 / 53.388 O b. 106.776 / 53.388 O c. None of the options listed is correct O d. More information is required to answer this question O e. 2776.990 / 27.528 Clear my choice

  • 5. Identify the following: k (number of groups or samples) N, (number of cases within each...

    5. Identify the following: k (number of groups or samples) N, (number of cases within each group or sample) N (total number of cases) df (between-groups degrees of freedom)= How is it calculated? df (within-groups degrees of freedom) How is it calculated? critical value for F (using the F table) SSB (Sum of Squares Between groups) = ssw (Sum of Squares Within groups) Mean square between= How is it calculated? Mean square within= How is it calculated? F ratio (test...

  • Complete the following ANOVA summary table using the appropriate formulas. Source Sum of Square df Mean...

    Complete the following ANOVA summary table using the appropriate formulas. Source Sum of Square df Mean Square Fobt Between groups 1,059 18 Within groups 3,702 167 N/A Total 4,761 185 N/A N/A Calculate the Mean Squared Between (bn)? Calculate the Mean Squared Within (wn)? Finally, calculate the F-Statistic or F-obtained?

  • ANOVA A study is designed to examine whether there is a difference in mean daily calcium intake a...

    ANOVA A study is designed to examine whether there is a difference in mean daily calcium intake among three groups of adults with normal bone density (Norm), adults with osteopenia (OstPNia) (a low bone density which may lead to osteoporosis) and adults with osteoporosis (OstPSis). A total of twenty-one adults at age 60 was recruited in the study (7 adults in each group). Each participant's daily calcium intake was measured based on reported food intake and supplements in milligrams. We...

  • Question Completion Status QUESTION 4 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 4...

    Question Completion Status QUESTION 4 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 4 108 27 32.66666667 B 4 96 24 13.33333333 4 120 30 56 ANOVA Source of Variation SS df F crit MS P-value F Treatments 72 36 1.058823529 0.386396621 4.256494729 Error 306 9 34 Total 378 11 Based on the Results above of Single Factor ANOVA: the MSTR O 36 O 378 O 72 O 34 Click Save and Submit to save and submit. Chick...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT