Question

A report included the following information on the heights (in.) for non-Hispanic white females. Sample Sample Std. Error Mean Size Age Mean 20-39 65.5 0.09 865 60 and older 938 63.7 0.11 (a) Calculate a confidence interval at confidence level approximately 95% for the difference between population mean height for the younger women and that for the older women. (Use H20-39 - Hs0 and older) Interpret the interval. O We are 95% confident that the true average height of younger women is greater than that of older women by an amount outside the confidence interval. O we are 95% confident that the true average height of younger women is less than that of older women by an amount within the confidence interval. O we cannot draw a conclusion from the given information. O We are 95% confident that the true average height of younger women is greater than that of older women by an amount within the confidence interval. (b) Let u, denote the population mean height for those aged 20-39 and H, denote the population mean height for those aged 60 and older. Interpret the hypotheses H,: H, - H, = 1 and H: u, - u, > 1. O The null hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is 1 inch higher than for older women. O The null hypothesis states that the true mean height for older women is 1 inch higher than for younger women. The alternative hypothesis states that the true mean height for older women is more than 1 inch higher than for younger women. O The null hypothesis states that the true mean height for younger women is 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women. O The null hypothesis states that the true mean height for older women is more than 1 inch higher than for younger women. The alternative hypothesis states that the true mean height for older women is 1 inch higher than for younger women. Carry out a test of these hypotheses at significance level 0.001. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) P-value =

Answer 1

Part a

Formula for confidence interval is given as below:

Confidence interval = (X1bar – X2bar) ± Z* sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

We use z distribution because sample sizes are larger enough.

From given information, we have

X1bar = 65.5

X2bar = 63.7

(X1bar – X2bar) = 65.5 – 63.7 = 1.8

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

n1 = 865

n2 = 938

Estimate for σ₁ = SE*sqrt(n) = 0.09*sqrt(865) = 2.6470

Estimate for σ₂ = SE*sqrt(n) = 0.11*sqrt(938) = 3.3689

SE = sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

SE = sqrt[(2.6470^2 / 865)+( 3.3689^2 / 938)]

SE = 0.1421

Confidence interval = (X1bar – X2bar) ± Z* sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

Confidence interval = 1.8 ± 1.96* 0.1421

Confidence interval = 1.8 ± 0.278516

Lower limit = 1.8 - 0.278516 = 1.521484

Upper limit = 1.8 + 0.278516 = 2.078516

Confidence interval = (1.5215, 2.0785)

We are 95% confident that the true average height of younger women is greater than that of older women by an amount within the confidence interval.

Part b

Null and alternative hypothesis:

The null hypothesis states that the true mean height for younger women is 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women.

Test statistic formula is given as below:

Z = (X1bar – X2bar) / sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

From part a, we have

(X1bar – X2bar) = 1.8

sqrt[(σ₁² / n₁)+(σ₂² / n₂)] = 0.1421

Z = 1.8/0.1421

Z = 12.66714

Test statistic = Z = 12.67

P-value = 0.0000

(by using z-table or excel)

P-value < α = 0.001

So, we reject the null hypothesis

There is sufficient evidence to conclude that the true mean height for younger women is more than 1 inch higher than for older women.