Question

A lumber company is making doors that are 2058.02058.0 millimeters tall. If the doors are too long they must be trimmed, and if the doors are too short they cannot be used. A sample of 1515 is made, and it is found that they have a mean of 2044.02044.0 millimeters with a standard deviation of 17.017.0. A level of significance of 0.050.05 will be used to determine if the doors are either too long or too short. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

Answer 1

Answer:

$t=\frac{2044-2058}{\frac{17}{\sqrt{15}}}=-3.190$    

$df=n-1=15-1=14$  

$p_v =P(t_{(14)}<-3.190)=0.0033$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly lower than 2058 mm at 5% of signficance.  

Step-by-step explanation:

Data given and notation    

$\bar X=2044$ represent the sample mean    

$s=17$ represent the sample standard deviation    

$n=15$ sample size    

$\mu_o =2058$ 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 mean is lower than 2058 mm, the system of hypothesis are :    

Null hypothesis:$\mu \geq 2058$    

Alternative hypothesis:$\mu < 2058$    

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:    

$t=\frac{2044-2058}{\frac{17}{\sqrt{15}}}=-3.190$    

P-value    

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

$df=n-1=15-1=14$  

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

$p_v =P(t_{(14)}<-3.190)=0.0033$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly lower than 2058 mm at 5% of signficance.