Question

The following table shows the average number of words understood by later-born and first-born one-year-old children from two random samples. Assume the population variances of the number of words understood are equal. Use the table to complete parts a through c. Later-Born First-Born Sample mean 437.4 556.0 Sample standard deviation 133.8 131.4 Sample size 13 13 a. Construct a 95% confidence interval to estimate the difference between the average number of words understood by first-born and later-born one-year-old children. The 95% confidence interval is ( CD) (Round to two decimal places as needed.) b. What conclusions can be made about the difference between the two populations? With 95% confidence, it can be said that the difference in the average number of words understood the endpoints of the confidence interval. The fact that this interval zero provides evidence that there a significant difference between the two populations. c. What assumptions need to be made in order to perform this procedure? A. Although the sample size in the second sample is large enough, it must be assumed that the distribution for the first sample is normally distributed. B. Although the sample size in the first sample is large enough, it must be assumed that the distribution for the second sample is normally distributed. C. It must be assumed that the distribution for both populations are normally distributed. It must also be assumed that the samples are independent. D. Since the sample sizes are large enough, no assumptions are needed.

Answer 1

a)

Degree of freedom, DF=   n1+n2-2 =    24              
t-critical value =    t α/2 =    2.0639   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    132.6054              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    52.0121              
margin of error, E = t*SE =    2.0639   *   52.0121   =   107.3478  
                      
difference of means =    x̅1-x̅2 =    556.0000   -   437.400   =   118.6000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    118.6000   -   107.3478   =   11.25
Interval Upper Limit=   (x̅1-x̅2) + E =    118.6000   +   107.3478   =   225.95

b)

Does not contain

there is a significant

c)

OC. It must be assumed that the distribution for both populations are normally distributed. It must also be assumed that the samples are independent.

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