The Waiting time for a Pizza order is supposed to be no more than 10 minutes. So

Question

The Waiting time for a Pizza order is supposed to be no more than 10 minutes. Some complaints have been received that the orders are in fact taking longer than the claimed 10 minutes. A batch of 81 orders was timed and found to have an average order time of 10.9 minutes and a sample standard deviation of (35/40) minutes. Prepare a hypothesis test to test whether the orders are taking longer than 10 minutes Test at 1% significance. Test whether the null hypothesis using critical value H_0: mu lessthanorequalto 10 should be accepted or rejected. Be sure to interpret your answer

Explanation / Answer

Having thet the null hypothesis is u>=10

Then, the alternative hypothesis is u>10.

We suppose than it is a normal distribution, so we use the formula

t*=x-u/(s/square root (n)),because we don't know the population deviation

Where x is sample's average = 10.9

u is population average = 10

s is satandar deviation of the sample = 35/40

and n is number of observation in the sample = 81

we substitude values in the formula and we have t= (10.9-10)/(35/40/sr(81))=9.25,with n-1 degrees of freedom (81-1)=80

We use de t distribution right-tail because we are proving that u>=10, that gives us t(80-1,1-alpha)=t(80,0.01)=2.373

If to>= t we accept ho, and if to<t we reject ho.

In this case we have to=9.25 and t=2.373 meaning that to>t, so we accept Ho.

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