Question

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar businesses reported the mean number of customers waiting to have their oil changed on Saturday morning is 3.6. Suppose the local oil changing business owner, wants to perform a hypothesis test. The null hypothesis is the population mean is 3.6 and the alternative hypothesis is the population mean is not equal to 3.6. The owner takes a random sample of 16 Saturday mornings during the past year and determines the sample mean is 4.2 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The level of significance is 0.05. The decision rule for this problem is to reject the null hypothesis if the observed "t" value is _______.

Answer 1

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X=4.2$ represent the sample mean  

$s=1.4$ represent the sample standard deviation  

n=16 represent the sample selected  

$\alpha=0.05$ significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:$\mu = 3.6$    

Alternative hypothesis:$\mu \neq 3.6$    

If we analyze the size for the sample is < 30 and we 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:    

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.