3. The annual salaries of workers at two companies, X andY, are to be compared.

Question

3. The annual salaries of workers at two companies, X andY, are to be compared. The salaries, in tens of thousands of dollars, at X andY, are denoted by x and y respectively. For a random sample of 40 workers in company X and a random sample of 50 workers in company Y the results are as follows. x-257.0 x2-|911 y 382.8 ?? 3150 The population mean salaries for X and r are denoted by ?? and ?, respectively. The population variances for salaries at x and Y can be assumed to be equal. Test, at the 5% significance level, whether ty is greater than (8 Marks] The width of a ? % confidence interval for ?,-ur is found to be 1.82. Find ii) 14 Marks] the value ofa.

Explanation / Answer

Here sample mean of x ; xbar = ?x/n = 257/40 = 6.425

sample Variance sx2 = 1/(n-1) [?x2 - (?x)2/n] = 6.661

sample standard deviation sx = 2.581

sample mean of y ; ybar = ?y/n = 382.8/50 = 7.656

sample Variance sy2 = 1/(n-1) [?y2 - (?y)2/n] = 4.4752

sample standard deviation sy = 2.1155

Here pooled standard deviation sp = sqrt[{(n1 -1)sx2 + (n2 -1)sy2 }/ (n1 + n2 -2)]

= 2.3332

Test statistic

t = (7.656 - 6.425)/[2.3332 * sqrt(1/40 + 1/50)]

t = 1.231/0.495

t = 2.487

Here dF = 88 and alpha = 0.05

tcritical =t88,0.05 = 1.6623

so here t > tcritical so we reject the null hypothesis

Margin of error = Critical test statistic * Standard error of mean

Standard error of mean = sp * sqrt(1/n1 + 1/n2) = 2.3332 * sqrt(1/40 + 1/50) = 0.4949

Width of confidence interval = 2 * margin of error

margin of error = 1.82/2 = 0.91

0.91 = tcritical * 0.4949

tcritical = 1.8386

Here dF = 88

so here alpha = 0.07

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