From candy to jewelry to flowers the average consumer was ex

From candy to jewelry to flowers, the average consumer was expected to spend $104.2 for Mother\'s Day in 2005, according to the Democrat & Chronicle article \"Mom\'s getting more this year\" (May 7, 2005). Local merchants thought this average was too high for their area. They contracted an agency to conduct a study. A random sample of 59consumers was taken at a local shopping mall the Saturday before Mother\'s Day and produced a sample mean amount of $94.67. If = $29.87, does the sample provide sufficient evidence to support the merchants\' claim at the .05 significance level?

(a) Find z. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion.

Reject the null hypothesis, there is not significant evidence to support the merchants\' claim.Reject the null hypothesis, there is significant evidence to support the merchants\' claim.     Fail to reject the null hypothesis, there is significant evidence to support the merchants\' claim.Fail to reject the null hypothesis, there is not significant evidence to support the merchants\' claim.

Solution

(a) z = (X-Mu)/sigma

     z=(104.2-94.67)/29.87

     z=0.31904

(b) By using STANDARD NORMAL DISTRIBUTION table we can calculate the p-value for z score

and p value for z=0.31904 is 0.6255

(c) Here p- value is 0.6255 and significance level is .05 so p -value is greater than significance indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis mean

there is significant evidence to support the merchants\' claim. if A small p-value (typically 0.05) indicates strong evidence against the null hypothesis or there is not significant evidence to support the merchants\' claim.Reject the null hypothesis.