Question

The quality control manager at a light bulb factory needs to estimate the average life of a large shipment of light bulbs. The process standard deviation is known to be 100 hours. A random sample of 64 light bulbs indicates a sample average of 350 hours. State the null and alternate hypothesis. At the 0.05 level of significance is there evidence that the average life is different from 385 hours. At the 0.10 level of significance is there evidence that the average life is greater than 385

Answer 1

Answer:

Null hypothesis:$\mu = 385$  

Alternative hypothesis:$\mu \neq 385$

$z=\frac{350-385}{\frac{100}{\sqrt{64}}}=-2.8$    

$p_v =2*P(z$  

At the 0.05 level of significance is there evidence that the average life is different from 385 hours

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Solution to the problem

Data given and notation  

$\bar X=350$ represent the sample mean

$\sigma=100$ represent the population standard deviation for the sample  

$n=64$ sample size  

$\mu_o =385$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 385, the system of hypothesis would be:  

Null hypothesis:$\mu = 385$  

Alternative hypothesis:$\mu \neq 385$  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{350-385}{\frac{100}{\sqrt{64}}}=-2.8$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z$  

At the 0.05 level of significance is there evidence that the average life is different from 385 hours