Question

A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. State the hypotheses to test whether the mean transaction time exceeds 60 seconds.

Answer 1

Answer:

Null hypothesis:$\mu \leq 60$  

Alternative hypothesis:$\mu > 60$  

Since we don't know the population deviation, 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:  

$t=\frac{67-60}{\frac{12}{\sqrt{16}}}=2.33$  

Step-by-step explanation:

Data given and notation  

$\bar X=67$ represent the sample mean    

$s=12$ represent the population standard deviation for the sample  

$n=16$ sample size  

$\mu_o =60$ represent the value that we want to test  

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

z 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 mean is higher than 60, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 60$  

Alternative hypothesis:$\mu > 60$  

Since we don't know the population deviation, 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:  

$t=\frac{67-60}{\frac{12}{\sqrt{16}}}=2.33$