Statistics check to see which answer is correct

1 Latoya Stevenson and Skye Keys Group Assignment- Independent (No response from other members) Date 07-26-2021

1.

sample proporGon is given by

The test staGsGc is given by

H0: p ≤ 0.6

H1: p > 0.6

n = 850

n o .  of people with a n t i bo dies = 540

P̂ = 540 850

= 0 . 6353

Z = P̂ − P0

P0(1 − P0 n

Z = 0.6353 − 0.60

0.60(1 − 0.60) 850

Z = 2.10077

2 Latoya Stevenson and Skye Keys Group Assignment- Independent (No response from other members) Date 07-26-2021

Since Z value is 2.10077 which is greater than the criGcal value 1.65, We reject the null hypothesis and conclude that there is significant evidence to accept the researcher’s claim that more than 60% of Americans have anGbodies for Coronavirus at 5% significance level.

2.

8

∑ i=1

Xdi = − 40

Pressure Support Before Treatment

Pressure Support AVer Treatment

Difference d

D2

20 17 -3 9

12 13 1 1

17 11 -6 36

10 6 -4 16

30 22 -8 64

23 15 -8 64

15 8 -7 49

5 0 -5 25

-40 264

t = ( ∑ D)/N

∑ D2 − ( ( ∑ D) 2

N )

(N − 1)(N )

t = −40/8

264 − ( ( − 40) 2

8 )

(8 − 1)(8)

t = − 4.6771

3 Latoya Stevenson and Skye Keys Group Assignment- Independent (No response from other members) Date 07-26-2021

Degrees of freedom

T- criGcal is ±3.499

Since calculated t value -4.6771 is outside the interval ±3.499 of the criGcal t value, we reject the null hypothesis and conclude that there staGsGcally significant evidence to accept the claim that there is no difference between the means at 0.01 significance level.

3. EXTRA CREDIT-

Step 1

The group means and overall mean

= n − 1 = 7

H0: μ1 = μ2 = μ3 = μ4 H1: Atlea t on e m ea n i s di f fer en t

X̄ = 1 N

N

∑ i=1

xi

X̄1 = 7.8 + 8.2 + … + 7

8 = 8.4

X̄2 = 7.6 + 8.1 + … + 6.1

8 = 7.16250

X̄3 = 6.8 + 6.5 + … + 8.1

8 = 7.26250

X̄4 = 9.2 + 8.1 + … + 7

8 = 7.23750

4 Latoya Stevenson and Skye Keys Group Assignment- Independent (No response from other members) Date 07-26-2021

The overall mean

Step 2: Calculate the Regression Sum of

vSquares (SSR)

Step 3: Calculate the Error Sum of Squares (SSE)

STEP 4: Total Sum of Squares

Step 5: Fill the ANOVA TABLE

X̄ = 240.5

32 = 7.515625

SSR = 8(8.4 − 7.515625)2 + 8(7.16250 − 7.515625)2 + 8(7.26250 − 7.515625)2 + 8(7.23750 − 7.515625)2 = 8.385938

SSE = 8

∑ i=1

(Xij − X̄j) 2

B AY A R E A = (7.8 − 8.4)2 + … + (7 − 8.4)2 = 10.22

VA L L E Y = (7.6 − 7.16250)2 + … + (6.1 − 7.16250)2 = 6.85875

MOUN TA I NS = (6.8 − 7.26250)2 + … + (8.1 − 7.26250)2 = 8.93875

DESERT = (9.2 − 7.23750)2 + … + (7 − 7.23750)2 = 11.75875

SSR = 37.816236

SST = SSR + SSE

SST = 8.385938 + 37.816236 = 46.202174

Source DF Sum of Square Mean Square F Sta1s1c P-value

Groups (between groups)

3 8.385938 2.795313 2.069713 0.126874

Error (within groups) 28 37.816236 1.350580

5 Latoya Stevenson and Skye Keys Group Assignment- Independent (No response from other members) Date 07-26-2021

F-criGcal is 2.95 at 0.05 level of significance and 3, 28 degrees of freedom.

Since F staGsGc from the ANOVA table is less than the F criGcal from F-distribuGon table, we fail to reject the null hypothesis and conclude that there is no enough evidence to accept the claim that at least one mean of working Gmes is different from others at 0.05 significance level.

Total 31 46.202174 1.490393