Question

An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it was determined that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group is suspicious of the claim and believes it is false. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W. What is the critical value with a level of significance of 0.05 and what is the test statistic and what is the p value?

Answer 1

Answer:

For this case the $\alpha=0.05$ and since we are conducting a right tailed test we need to find a critical value on the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{cric}= 1.64$

$z=\frac{257.3-250}{\frac{15}{\sqrt{20}}}=2.176$    

$p_v =P(z>2.176)=0.0148$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly higher than 250 W at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=257.3$ represent the sample mean

$\sigma=15$ represent the population standard deviation

$n=20$ sample size  

$\mu_o =250$ represent the value that we want to test

$\alpha=0.05$ 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 check if the true mean is higher than 250, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 250$  

Alternative hypothesis:$\mu > 250$  

If we analyze the size for the sample is < 30 but 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".  

Critical value

For this case the $\alpha=0.05$ and since we are conducting a right tailed test we need to find a critical value on the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{cric}= 1.64$

Calculate the statistic

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

$z=\frac{257.3-250}{\frac{15}{\sqrt{20}}}=2.176$    

P-value

Since is a right tailed test the p value would be:  

$p_v =P(z>2.176)=0.0148$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly higher than 250 W at 5% of signficance.