Question

A lumber company is making doors that are 2058.0 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 22 is made, and it is found that they have a mean of 2045.0 millimeters with a standard deviation of 13.0. A level of significance of 0.1 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:

$t=\frac{2045-2058}{\frac{13}{\sqrt{22}}}=-4.69$      

The degrees of freedom are given by:

$df=n-1=22-1=21$  

And the p value would be given by:

$p_v =2*P(t_{21}<-4.69)=0.000125$  

Since the p value is a very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2058 mm at the significance level of 0.1 (10%) given

Step-by-step explanation:

Information given

$\bar X=2045$ represent the sample mean      

$s=13$ represent the standard deviation

$n=22$ sample size      

$\mu_o =2058$ represent the value to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to cehck if the true mean for this case is equal to 2058 or not, the system of hypothesis would be:      

Null hypothesis:$\mu = 2058$      

Alternative hypothesis:$\mu \neq 2058$      

The statistic for this case is given by:

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

And replacing we got:

$t=\frac{2045-2058}{\frac{13}{\sqrt{22}}}=-4.69$      

The degrees of freedom are given by:

$df=n-1=22-1=21$  

And the p value would be given by:

$p_v =2*P(t_{21}<-4.69)=0.000125$  

Since the p value is a very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2058 mm at the significance level of 0.1 (10%) given