Answer to Question #229162 in Statistics and Probability for sanzz

Question

 A shopkeeper claims that the average life of a CFL Bulb is 1600 hours. To check this claim, a researcher takes a sample of 100 CFL bulbs of the same make randomly and finds mean lifetime of 1570 hours with standard deviation of 120 hours. Is the claim acceptable at 5% level of significance ? [Given that Z= ± 1.96]

Answer 1

Solution:

n=100, $\bar X=1570, \sigma=120$

We test the hypotheses ;

$H_0:\mu=1600$

$H_1:\mu≠1600$

Since the sample size is greater than 30 and the population standard deviation is known we use the z test.

$Z=\dfrac{X-\mu} {\dfrac{\sigma} {\sqrt n}}$

=$\dfrac{1570-1600}{\dfrac{120}{\sqrt{100}}}$

=-2.5

We take a level of significance of 5%.

The corresponding critical value is;

$Z_{\alpha/2}=1.96$

If the test statistic is greater than the critical value, we reject the null hypothesis in favor of the alternative hypothesis.

|-2.5|>1.96 or 2.5>1.96

We reject the null hypothesis hypothesis in favor of the alternative hypothesis.

We are 95% confident that the population mean lifetime of bulbs is not equal to 1600 hours.