Question

according to a report done by s & j power, the mean lifetime of the light bulbs it manufactures is 55 months. a researcher for a consumer advocate group tests this by selecting 100 bulbs at random. for the bulbs in the sample, the mean lifetime is 56 months. it is known that the population standard - deviation of the lifetimes is 6 months. can we conclude, at the 0.01 level of significance, that the population mean lifetime, μ, of light bulbs made by this manufacturer differs from 55 months? (a) state the null hypothesis h0 and the alternative hypothesis h1. h0 : μ = 55 h1 : μ ≠ 55 (b) determine the type of test statistic to use. z (c) find the value of the test statistic. (round to three or more decimal places.) (d) find the p - value. (round to three or more decimal places.) (e) can we conclude that the population mean lifetime of light bulbs made by this manufacturer differs from 55 months? o yes o no

Explanation:

Step1: Identify hypotheses

$H_0:\mu = 55$, $H_1:\mu
eq55$

Step2: Assume we have sample - mean $\bar{x}$, sample - standard deviation $s$ and sample size $n$. The formula for the z - test statistic (if population standard deviation $\sigma$ is known) or t - test statistic (if $\sigma$ is unknown) is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ (z - test) or $t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ (t - test). Let's assume we are using a z - test and we have $\bar{x} = 56$, $\sigma = 5$, $n = 100$. Then $z=\frac{56 - 55}{\frac{5}{\sqrt{100}}}=\frac{1}{\frac{5}{10}} = 2$.

Step3: Calculate p - value

Since it's a two - tailed test, the p - value is $2\times(1 - P(Z<|z|))$. If $z = 2$, then $P(Z < 2)=0.9772$, and the p - value is $2\times(1 - 0.9772)=2\times0.0228 = 0.0456$.

Step4: Make a decision

The significance level $\alpha=0.01$. Since $p - value=0.0456>0.01$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 55$, $H_1:\mu
eq55$
(b) Two - tailed test
(c) Assume $z = 2$ (example value based on assumed data above)
(d) $p - value = 0.0456$
(e) No