Question

for following questions please 1. find critcal value and 2. compute test value. clearly idenify. thanks so much! your help is appreciated!

UR Is there a significant difference at a = 0.05 in the mean heights in feet of waterfalls in Europe and the ones in Asia? The data are shown. Source:World Almanac. Use the critical value method with tables. Asia 470 900 1246 Europe 345 820 1385 984 614 1137 350 722 964 320 1904 722 Download data Use , for the mean height of waterfalls in Europe. Assume the variables are normally distributed and the variances are unequal. 1 Part 1 State the hypotheses and identify the claim with the correct hypothesis. H: H2 not claim HH2 - 2 claim This hypothesis test is a two-tailed test. Part 2 out of 5 Pony Tutorial help Plugin not installed or missing? Show More 31 MACBOOK Pro

Answer 1

Q1:

For Europe :

∑x = 6150

∑x² = 6261322

n1 = 7

Mean , x̅1 = Ʃx/n = 6150/7 = 878.5714

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(6261322-(6150)²/7)/(7-1)] = 378.1771

For Asia :

∑x = 6733

∑x² = 7491745

n2 = 8

Mean , x̅2 = Ʃx/n = 6733/8 = 841.6250

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(7491745-(6733)²/8)/(8-1)] = 510.6136

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 12.7032 = 13

Critical value, t_c = T.INV.2T(0.05, 13) = 2.160

Reject Ho if t < -2.16 or if t > 2.16

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (878.5714 - 841.625)/√(378.1771²/7 + 510.6136²/8) = 0.160

Decision:

Do not reject the null hypothesis

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Q2:

For PGA :

∑x = 41178

∑x² = 3.1E+08

n1 = 9

Mean , x̅1 = Ʃx/n = 41178/9 = 4575.3333

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(312099292-(41178)²/9)/(9-1)] = 3932.1784

For LPGA :

∑x = 10836

∑x² = 3.2E+07

n2 = 9

Mean , x̅2 = Ʃx/n = 10836/9 = 1204.0000

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(32091002-(10836)²/9)/(9-1)] = 1542.9055

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 10.4063 = 10

Critical value, t_c = T.INV.2T(0.01, 10) = 3.169

Reject Ho if t < -3.169 or if t > 3.169

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (4575.3333 - 1204)/√(3932.1784²/9 + 1542.9055²/9) = 2.394

Decision:

Do not reject the null hypothesis

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Q3:

Sample 1:

Sample Variance using excel function VAR.S(), s₁² = 11113.8

Sample size, n₁ = 16

Sample 2:

Sample Variance excel function VAR.S(), s₂² = 14780.86

Sample size, n₂ = 16

Null and alternative hypothesis:

Hₒ : σ₁² = σ₂²

H₁ : σ₁² < σ₂²

Test statistic:

F = s₁² / s₂² = 11113.8 / 14780.8625 = 0.752

Degree of freedom:

df₁ = n₁-1 = 15

df₂ = n₂-1 = 15

Critical value(s):

Lower tailed critical value, FL = F.INV(0.05, 15, 15) = 0.416

Conclusion:

We fail to reject the null hypothesis.