Question

Does training in study skills make a difference for student success? 8 students received training (M₁= 85, SS₁= 100) and 8 students did not (M₂= 80, SS₂= 124). Test using α = .05  

Which is the null hypothesis, H₀?

µ₁ = 0

µ₁- µ₂ ≠ 0

µ₁- µ₂ = 0

µ₁ ≠ 0

The degrees of freedom (total) (df) =

18

14

16

10

With alpha = .05, (two tail, nondirectional test), the critical t score boundaries =

+/- 2.145

+/- 2.179

+/- 2.120

+/- 2.776

s²_p= (pooled variance) =

20

16

18

12

s_{(M1-M2) =}

8

2

4

1

t = (test statistic) =

+1.75

+3.00

+1.25

+2.50

What is the correct result based on the data?

Reject H_o, result is Significant

Reject H_o, result is NOT Significant

Retain H_o, result is NOT Significant

Retain H_o,result is Significant

What type of error could you be making?

Type I

Type II

Does training in study skills make a difference?

No

Yes

The correct APA format for the result would include

p > .05

p < .05

I know this is a lot but it's all one problem. You dont have to answer all if you cannot, but it would be so much appreciated.

Answer 1

ni -8 XI - Given : 85 SS 100 n2-8 X2 = 80 SS2 = 124 Ho: Mi-42=0 Hi: li-42 € 0 df = nitn2-2 = 8 + 8-2 14 critical value at x = 0.05, df=14 12.145 tcritical df to.025, 14 (from t-table) pooled variance ( sº) Så sSi+$52 nitha-2 2 100 t124 ast 8 +8 2 Sp 224 14 sp 16 Test statistic ta 71-72 Spree t ni กว + 85-80 16 + 16 8 + 은 t = 2.50

since, iti > tcritical, Wale critical, we reject Ho and conclude that result is significant type of Error = Type - training in study make difference correct APA format is . p < 0.05