Question

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. 

Answer 1

$H_0:u_a=u_b$

$H_0:u_a>u_b$

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