Question

Question 3 of 10 (4 points) | Attempt 1 of 1 | View question in a popup | © 2h 55m Remaining 11.1 Section Exercise 16 (p-value, calc) ol. Baby weights: Following are weights in pounds for random samples of 17 newborn baby boys and baby girls born in Denver in 2011. Box plots indicate that the samples come from populations that are approximately normal. Let u, denote the mean weight of boys. Can you conclude that the mean weight boys is greater than the mean weight of girls? Use the Q = 0.05 level and the P-value method with the TI-84 Plus calculator. 7.3 6.3 7.4 Boys 8.5 7.7 8.3 7.5 7.8 6.4 7.8 8.4 8.6 6.9 7.4 6.9 7.7 8.1 Girls 5.7 8.5 7.2 6.9 8.2 6.5 7.5 6.7 6.6 6.4 6.3 5.9 7.2 8.1 8.2 6.7 6.2 Send data to Excel Part: 0 / 4 Part 1 of 4 Stato the null and alternato bunsthacoe Next Part Submit Assignment

Answer 1

For Boys :

∑x = 129

∑x² = 986.46

n1 = 17

Mean , x̅1 = Ʃx/n = 129/17 = 7.5882

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(986.46-(129)²/17)/(17-1)] = 0.6882

For Girls :

∑x = 118.8

∑x² = 841.86

n2 = 17

Mean , x̅2 = Ʃx/n = 118.8/17 = 6.9882

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(841.86-(118.8)²/17)/(17-1)] = 0.8536

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Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 > µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((17-1)*0.6882² + (17-1)*0.8536²) / (17+17-2) = 0.6011

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (7.5882 - 6.9882) / √(0.6011*(1/17 + 1/17)) = 2.2562

df = n1+n2-2 = 32

p-value = T.DIST.RT(2.2562, 32) = 0.0155

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that  the mean weight of boys is greater than the mean weight of girls at 0.05 significance level.