The label on 1-gallon can of paint states that the amount of paint in the can Is sufficient to paint at least 400 square feet (on average). Suppose the amount of coverage Is approximately normally distributed, and the overall standard deviation of the amount of coverage is 27 square feet. A random sample of 16 cans yields the sample mean amount of coverage of 390 square feet. Test the claim on the label at a 5% level of significance. Enter the test statistic in the first box and your decision in the second box (0 if you reject and 1 if you don't (i.e if you accept) Find the p value of the test in part (a)

Answer:[tex]z=\frac{390-400}{\frac{27}{\sqrt{16}}}=-1.481[/tex]        [tex]p_v =P(z<-1.481)=0.0693[/tex]   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 FAIL reject the null hypothesis, so is not enough evidence to conclude that the sample mean is less then 400 at 5% of significance. Step-by-step explanation:Data given and notation        [tex]\bar X=390[/tex] represent the mean for the sample    [tex]\sigma=27[/tex] represent the standard deviation for the population        [tex]n=16[/tex] sample size        [tex]\mu_o =400[/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 400[/tex]        Alternative hypothesis:[tex]\mu < 400[/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{390-400}{\frac{27}{\sqrt{16}}}=-1.481[/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<-1.481)=0.0693[/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 FAIL reject the null hypothesis, so is not enough evidence to conclude that the sample mean is less then 400 at 5% of significance.