Question

A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 732 hours. A random sample of 28 light bulbs has a mean life of 704 hours. Assume the population is normally distributed and the population standard deviation is 65 hours. At alphaequals0.05​, do you have enough evidence to reject the​ manufacturer's claim?

Answer 1

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ ≥ 732

For the alternative hypothesis,

µ < 732

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = lifetime of the bulb

µ = mean lifetime

σ = standard deviation

n = number of samples

From the information given,

µ = 732

x = 704 hours

σ = 65 hours

n = 28

z = (704 - 732)/(65/√28) = - 2.28

Looking at the normal distribution table, the probability corresponding to the z score is 0.011

Since alpha, 0.05 > than the p value, 0.011, then we would reject the null hypothesis. Therefore, At a 5% level of significance, there is enough evidence to reject the​ manufacturer's claim

Answer 2

Answer:

$z=\frac{704-732}{\frac{65}{\sqrt{28}}}=-2.279$      

$p_v =P(z$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and the the true mean is significantly lower than 732 hours so we have enough evidence to reject the claim

Step-by-step explanation:

Data given and notation      

$\bar X=704$ represent the sample mean

$\sigma=65$ represent the standard deviation for the population      

$n=28$ sample size      

$\mu_o =732$ represent the value that we want to test    

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

t 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 determine if the true mean is at least 732 or no, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 732$      

Alternative hypothesis:$\mu < 732$      

We know the population deviation, so for this case 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:      

$z=\frac{704-732}{\frac{65}{\sqrt{28}}}=-2.279$      

Calculate the P-value      

Since is a one-side lower test the p value would be:      

$p_v =P(z$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and the the true mean is significantly lower than 732 hours so we have enough evidence to reject the claim