Question

A revenue department is under orders to reduce the time small business owners spend filling outpension form ABC-5500. Previously the average time spent on the form was 5.7 hours. In orde r to test whether the time to fill out the form has been reduced, a sample of 86 small business owners who annually complete the form was randomly chosen, and their completion times recorded. The mean completion time for ABC -5500 form was 5.5 hours with a standard deviation of 2.3 hours. In order to test that the time to complete the form has been reduced, state the appropriate null and alternative hypotheses.Calculate the test statistic.

Answer 1

Answer:

Null hypothesis:$\mu \geq 5.7$    

Alternative hypothesis:$\mu < 5.7$    

$t=\frac{5.5-5.7}{\frac{2.3}{\sqrt{86}}}=-0.806$    

Step by step explanation

1) Data given and notation    

$\bar X=5.5$ represent the mean completion time for ABC -5500 form for the sample    

$s=2.3$ represent the standard deviation for the sample    

$n=86$ sample size    

$\mu_o =5.7$ represent the value that we want to test  

$\alpha$ 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 determine if the mean completion time for ABC -5500  is less than 5.7, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 5.7$    

Alternative hypothesis:$\mu < 5.7$    

We don't know the population deviation, so for this case 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:    

$t=\frac{5.5-5.7}{\frac{2.3}{\sqrt{86}}}=-0.806$    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=86-1=85$

Since is a one-side lower test the p value would be:    

$p_v =P(t_{85}<-0.86)=0.196$

Conclusion    

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean completion time for ABC -5500 is not significantly lower than 5.7, the previous average.