Answer: Test statistic is equal to 2.40 with 2 degrees of freedom. Accept null h

Question

Answer: Test statistic is equal to 2.40 with 2 degrees of freedom. Accept null hypothesis at the 0.10, 0.05, 0.01 level. A research team took a random sample of 5 observations from a normally distributed random variable Y and observed that ys-3442 and s,-129.7 where y, was the average of the five observations sampled from I and s was the unbiased estimate of var()(ie, the divisor in the variance was n-1). A second research team took a random sample of 4 observations from a normally distributed random variable X and observed that x, = 184.6 and s-1456. where x, was the average of the four observations sampled from l and s was the unbiased estimate of var(X) (i.e., the divisor in the variance was n-1) Calculate the 95% confidence interval for E(X)-E(F) using the pooled variance estimator. This problem is worth 40 points 2.

Explanation / Answer

Concept Back-up

Given X ~ N(1, 12), Y ~ N(2, 22), 1 = 2 = , say unknown, n1 n2

100(1 - ) % Confidence Interval for (1 - 2) is: (Xbar – Ybar) ± {( tn1 + n2 – 2, /2)(s){(1/n1) + (1/n2)}, where Xbar and Ybar are sample means, s = pooled sample estimate of given by s = sqrt[{(n1– 1)s12+ (n2– 1)s22)}/(n1+ n2– 2)]; s1, s2are respective sample standard deviations; tn1 + n2 – 2, /2= upper (/2) percent point of t-distribution with degrees of freedom = n1+ n2– 2; and n1, n2are sample sizes.

Now, to work out the solution,

Given Xbar = 184.6, Ybar = 344.2, n1 = 4, s12 = 145.6, n2 = 5, s22 = 129.7, = 5%,

s2 = {(3 x 145.6) + (4 x 129.7)}/7

= (518.4 + 436.8)/7

= 955.2/7

= 136.4571

So, s = sqrt(136.4571) = 11.68

tn1 + n2 – 2, /2 = t7, 0.025 = 2.365 [from standard t-distribution table]

{(1/n1) + (1/n2)} = (0.25 + 0.2) = 0.45 = 0.6708

95% Confidence Interval for {E(X) - E(Y} is:

(184.6 - 344.2) ± (2.365 x 11.68 x 0.6708)

= - 159.6 ± 18.53

= (- 178.13, - 141.07) ANSWER

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