The mean lifetime of 30 bulbs produced by Company A is 500 hours and the mean lifetime of 35 bulbs produced by Company B is 492 hours. If the standard deviation of all bulbs produced by Company A is 10 hours and the standard deviation of all bulbs produced by Company B is 15 hours, test at 1% significance level that the mean lifetime of bulbs produced by Company A is better than of Company B.
"H_0:u_a=u_b"
"H_0:u_a>u_b"
where "u_a, u_b" - population mean lifetime of bulbs A and B respectively
The test statistic is calculated next way
"K={\\frac {x_a-x_b} {\\sqrt{{\\frac {\\sigma^2_x} n}+{\\frac {\\sigma^2_y} m}}}}" , where "x_a,x_b" - sample means of bulbs A, B respectively, "\\sigma^2_x,\\sigma^2_y" - sample standard deviations of bulbs A, B respectively, n and m - sample sizes of bulbs A and B respectively
In the given case we have
"K={\\frac {500-492} {\\sqrt{{\\frac {10^2} {30}}+{\\frac {15^2} {35}}}}}\\approx2.56"
Since population standard deviations is known, then it is appropriate to use Z-value as critical value. According to tha form of the alternative hypothesis, we should run one-tailed test, then
"P(Z>Cr)=1-\\alpha=0.99\\implies Cr=2.33" , Cr - critical value, "\\alpha" - level of significance
we receive that K > Cr, so there is enough statistical evidence to reject null hypothesis at 1% significance level, so we should conclude that lifetime of bulbs A is better than B
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