Question

The mean waiting time at a drive-through of a fast-food restaurant from the time the order is received is 84 seconds. A manager devises a new drive-through that he believes will decrease wait time. He initiates the new system at his restaurant and measures the wait time for 100 randomly selected orders. The sample mean is 81 seconds with a standard deviation of 3.6 seconds. Assume that wait times are normally distributed. The manager conducts a hypothesis test to determine if the new system decreases wait time. (alpha=0.05)
What is the standard score?

Answer 1

Answer:

$t=\frac{81-84}{\frac{3.6}{\sqrt{100}}}=-8.33$    

And the standard score for this case is $-8.33$

Step-by-step explanation:

Data given and notation  

$\bar X=81$ represent the sample mean

$s=3.6$ represent the sample standard deviation

$n=100$ sample size  

$\mu_o =84$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the ture mean is lower than 84, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 84$  

Alternative hypothesis:$\mu < 84$  

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{81-84}{\frac{3.6}{\sqrt{100}}}=-8.33$    

And the standard score for this case is $-8.33$