Question 358596
Given
{{{mu=3.55}}} and {{{sigma=1.1}}}
sample n=100 and sample mean xbar=3.18
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hypothesis 
Ho: {{{mu=3.55}}}
Ha: {{{mu<3.55}}}  lower tail test
Note: I dont see where you provide the signifance level to perform the test
I'll assume {{{alpha=0.05}}}
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a. At the level of significance, using the critical value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes? 

we dont know the population distribution.  But since we are testing sample averages and the sample size is >30 we can rely on the "Central Limit Theorem" to assume a normal distribution for the sample averages.
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furthermore since we are told the population standard deviation {{{sigma=1.1}}}
we know that we can use a critical value based on Z
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lower tail critical value is Z at a tail probability of {{{alpha=0.05}}}
Z=-1.645
What this means is that sample standardized values as extreme as -1.645 can be expected by pure chance if the null hypothesis {{{mu=3.55}}} is assumed
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Compute the test statistic  {{{Z=(xbar-mu)/(sigma/sqrt(n))}}} or Z=(3.18-3.55)/(1.1/sqrt(100))=-3.36
Since our test statistic (our evidence for the null) is more extreme than the critical value of Z=-1.645 then we reject the null.  This cannot happen by pure chance.  
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b. At the level of significance, using the p-value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes? 
pvalue =P(Xbar<3.18) given the assumption of the null {{{mu=3.55}}}
P(Xbar<3.18)=P(Z<-3.36)=very small (less than 0.005) So we conclude that a value of Xbar as extreme as 3.18 cannot happen by pure chance under the null, so we reject the null hypothesis and conclude that evidence suggest {{{mu<3.55}}}