Question

PLEASE INCLUDE GRAPH!!! and please answer correctly

3 2004 Hypothesis test for 1 mean Page 406-413 A fast food restaurant claims that the mean waiting time is greater than random sample of 50 customers revealed a sample mean waiting time of 2.0 the mean waiting time is greater than 3.2 minutes (original claim). A deviation of 0.486 minutes. revealed a sample mean waiting time of 2.8 minutes with a sample standard Test the restaurants claim at a a= 0.05 level of si the population has a normally distribution and the population standar nts claim at a = 0.05 level of significance. Assume that Traditional method to test the restaurant's claim Use the guidelines in textbook tribution and the population standard deviation is unknown. Use the (with graph) and number each step. (Hint: Use t-test) 15 points total. graph 15 Po

Answer 1

To perform the right-tail hypothesis test, we use the six-step method. Step1: The null and alternative hypotheses are as follows. H,:4=32 (The population mean is equal to 3.2) versus H:4>3.2 The population mean is greater than 3.2) Step 2: We have selected a sample of n = 50. We decide to use a = 0.05. Step3: Because o is unknown, you use the t - distribution and the t-test statistic. We must assume that the population mean are normally distributed - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step4: For a given sample size, n, the test statistic t follows at distribution with n-1 degrees of freedom. The critical values of the t distribution with 50-1 = 49 degrees of freedom. The table of t test is obtained by using the Excel2010 function, = tinv(probability degrees of freedom) timv(0.05,49) = 1.676550893 10.05.09 = 1.676550893 --------------- -------------------------- --- Step5: The sample statistics: The sample mean is X = 2.8 minutes. The standard deviation is s= 0.486 minutes The sample size is n = 50. ----- ------ -------- -------- The t-test statistic is given by ---------- ------ -- X- _2.8-3.2 0.486 150 -0.4 = 0.486 7.071068 -0.4 0.068731 =-5.81981 t = -5.82 Step 6: The P-value approach: P(t> )=(1-P(t <to)) =(1-P(1<<,)) =(1-P{{<-5.82)) =(1-0.999999779) =0.000000221 P-value 0.000000221 <a= 0.05 We reject the null hypothesis H. ------------------------------------- The graph of the test statistic and the corresponding p-value, critical value of the t-distribution:

Distribution Plot Distribution Plot T, df=49 P-value = 0.0000 Density 0.05 0 Critical value=1.677 t=5.82

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Step7: Conclusion: Since, the value of the probability is P-value = 0.000000221 <a = 0.05 We reject the null hypothesis Ho. Since, the test statistic is bl=5.82 > 60 05:49 = 1.676550893 We reject the null hypothesis Ho There is sufficient evidence to conclude that the mean waiting time is greater than 3.2 minutes at the 5% level of significance.