7. The dletribbrtion of theF rato Aa Lek\'s say that researcherconducts.sEusly w

Question

7. The dletribbrtion of theF rato Aa Lek's say that researcherconducts.sEusly with 4 groups, each with 6 partidperts FVl inthe degrees of hedorn In the following ANOVA table df HS 23 29 3.95 5.89 Source within Total 23 use the following Distributionstool to findthe boundary for th crtical rogon at ".05 and -.01 T0osethe ter to find the crit al F value, set both the numerator and the denomiator degrees of freedom; this w show you the appropriate F distribution. Move the orange Hoe until the aree in the tail is equivalent to the apha leve you are imvestigating Numeator Degrees of Freed2 Atthe a .05 level of signficana, te boundary ofthe critical rmjon for this ANOVA ias O 3.009 O 4938 O 4.718 O 3.098 Atthe ·01 level of sign iana, the boundary ofthe cticat region for this ANOVAis: O 4.718 0 3.098 03.009. O 4.938 wow use the tool to evaluate the F-ratio. (Hinti Select the icon with one orange tine and one purple line. You can set the orange tine at the orical boundary and move the purple line to the F-ratio ia the ANQVA table given To use the tool to find the F-ratio, set hoth the aurserator aad the denom inator degrees of freedom, this m show you the appropriate F-distribution. Move the orange ine snbi the area in the taul is equtvalent to the alpha teret you nul: hypothesis can be rejected O The nudl hypothesis cannot be rejected

Explanation / Answer

Answer:

df

MS

F

Between

3

23.29

3.95

within

20

5.89

Total

23

0.05 level , critical value = 3.098

0.01 level , critical value = 4.938

At 0.05 level, the null hypothesis can be rejected.

At 0.01 level, the null hypothesis can not be rejected.

df

MS

F

Between

3

23.29

3.95

within

20

5.89

Total

23

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.