Question

A report included the following information on the heights (in.) for non-Hispanic white females. Age Sample sample sitd. Error Size Mean Mean 09 0 3.9 0.09 937 62.4 0.11 20-39 60 and older (a) Calculate a confidence interval at confidence level approximately 95% for the difference between population mean helight for the younger women and that for the older women. (Use 20-39" so and older Interpret the interval We cannot draw a conclusion from the given information. 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. 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 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 , denote the population mean height for those aged 20-39 and denote the population mean height for those aged 60 and older. Interpret the hypotheses Wol H = 1 and 21 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. 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. 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. 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. carry out a test of these hypotheses at significance level 0.001. Calculate the test statistic and determine the p value. (Round your test statute to two decimal places and your value to four decimal places.) PM MIQH WAS

Answer 1

a)

Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.960   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.1421          
margin of error, E = Z*SE =    1.960   *   0.142   =   0.2786
                  
difference of means = x̅1 - x̅2 =    63.9   -   62.4   =   1.500
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    1.500   -   0.279   =   1.2214
Interval Upper Limit= (x̅1 - x̅2) + E =    1.500   +   0.279   =   1.7786

answer: option D)

b) option A)

difference in sample means = x̅1 - x̅2 =    63.9   -   62.4   =   1.5
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.1421          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    1.5   /   0.1421   =   3.52
                  
  
p-value =        0.0002 [excel function =NORMSDIST(z)]      

c) option d)

d) option C)