Answer :
Answer:
Critical value zc=1.64
[tex]z=\frac{257.3-250}{\frac{15}{\sqrt{20}}}=2.176[/tex]
[tex]p_v =P(z>2.176)=0.0147[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=257.3[/tex] represent the mean for the sample
[tex]\sigma=15[/tex] represent the standard deviation for the population
[tex]n=20[/tex] sample size
[tex]\mu_o =250[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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 compact microwave oven consumes a mean of no more than 250 W:
Null hypothesis:[tex]\mu \geq 250[/tex]
Alternative hypothesis:[tex]\mu > 250[/tex]
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:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (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:
[tex]z=\frac{257.3-250}{\frac{15}{\sqrt{20}}}=2.176[/tex]
Calculate the critical value
The critical value for this case would be :
[tex]P(Z>a)=0.05[/tex]
The value of a that satisfy this on the normal standard distribution is a=1.64 and would be the critical value on this case zc=1.64.
Calculate the P-value
Since is a one-side upper test the p value would be:
[tex]p_v =P(z>2.176)=0.0147[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the the actual mean of comsumption for the compact microwave oven seems to be higher than 250W.