Q:

# For the hypothesis test H0: μ = 10 against H1: μ &gt;10 and variance known, calculate the Pvalue for each of the following test statistics.(a) z0 = 2.05 (b) z0 = −1.84 (c) z0 = 0.4

Accepted Solution

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Answer:a) $$p_v =P(Z>2.05)=1-P(z<2.05)=1-0.980=0.0202$$b) $$p_v =P(Z>-1.84)=1-P(z<-1.84)=1-0.0330=0.967$$c) $$p_v =P(Z>0.4)=1-P(z<0.4)=1-0.655=0.345$$Step-by-step explanation:Some previous conceptsThe p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"HypothesisNull hypothesis: $$\mu=10$$Alternative hypothesis: $$\mu >10$$If the random variable is distributed like this: $$X \sim N(\mu,\sigma)$$We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:$$z_o=\frac{\bar X -\mu}{\sigma}$$Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:Part a$$p_v =P(Z>2.05)=1-P(z<2.05)=1-0.980=0.0202$$And in order to find the answer using excel we can use the following code:"=1-NORM.DIST(2.05,0,1,TRUE)"Part b$$p_v =P(Z>-1.84)=1-P(z<-1.84)=1-0.0330=0.967$$And in order to find the answer using excel we can use the following code:"=1-NORM.DIST(-1.84,0,1,TRUE)"Part c$$p_v =P(Z>0.4)=1-P(z<0.4)=1-0.655=0.345$$And in order to find the answer using excel we can use the following code:"=1-NORM.DIST(0.4,0,1,TRUE)"ConclusionsIf we use a reference value for the significance, let's say $$\alpha=0.05$$. For part a the $$p_v<\alpha$$ so then we can reject the null hypothesis at this significance level.For part b the $$p_v>\alpha$$ so then we FAIL to reject the null hypothesis at this significance level.For part c the $$p_v>\alpha$$ so again we FAIL to reject the null hypothesis at this significance level.