Question

Suppose we had the following summary statistics from two different, independent populations, both with variances equal to σ.

1. Population 1: ¯x1= 126, s1= 8.062, n1= 5
2. Population 2: ¯x2= 162.75, s2 = 3.5, n2 = 4

We want to find a 99% confidence interval for μ2−μ1. To do this, answer the below questions.

Suppose we had the following summary statistics from two different, independent populations, both with variances equal to o: 1. Population 1: Ti = 126, $i = 8.062, ni = 5 2. Population 2: 22 = 162.75, S2 = 3.5, n2 = 4 We want to find a 99% confidence interval for H2 – M1. To do this, answer the below questions. a. Can we assume equal variances or not? No, we cannot assume equal variances. Yes, we can assume equal variances. b. The pooled standard deviation is: s, = 6.511 Round to 3 decimal places. c. The standard error is: SE= * Hint Round to 3 decimal places. d. What is the degrees of freedom associated with this problem? (Round down to the nearest whole number.) The critical value from the distribution for a confidence interval of 99% is: **= * Use Technology Round to 3 decimal places. e. Using the the answers from above (all rounded to 3 decimal places), calculate by hand a 99% confidence interval for 42 - Mi. *< Select an answer A) * *. Hint Round to 3 decimal places. Answer: Yes, we can assume equal variances. Answer: 6.511 Answer: 4.368 Answer: 7 Answer: 3.500 Answer: 21.46 Answer: u_2-u_1 Answer: 52.04

Answer 1

(a) Yes, we can use pooled variance

(b) The pooled Standard Deviation

$S_p^2 = \frac{(n1-1)*s1^2+(n2-1)*s2^2}{n1+n2-2} = \frac{4*8.064^2+3*3.5^2}{5+4-2} = 42.3905$

Sp = 6.511

(c) The Standard Error (SE)

$\sqrt{S_p^2*(\frac{1}{n1}+\frac{1}{n2})} = \sqrt{42.3905*(\frac{1}{5}+\frac{1}{4})} = 4.368$

(d) Degrees of freedom = n1 + n2 - 2 = 5 + 4 -2 = 7

(e) The critical value, $t^*$ = 3.5

(f)

Lower Limit = (162.75 - 126) - [3.5 * 4.368] = 36.75 - 15.29 = 21.46

Upper Limit = (162.75 - 126) + [3.5 * 4.368] = 36.75 + 15.29 = 52.04

Therefore 21.46 < $\mu_1 - \mu_2$ < 52.04