Question

vlumber company is making boards that are 2944.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A sample of 25 is made, and it is found that they have a mean of 2941.0 millimeters with a standard deviation of 12.0. A level of significance of 0.1 will be used to determine if the boards 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:

$t=\frac{2941-2944}{\frac{12}{\sqrt{25}}}=-1.250$      

The degrees of freedom are given by:

$df=n-1=25-1=24$  

The p value would be given by:

$p_v =2*P(t_{24}<-1.250)=0.223$  

Since the p value is higher than 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2944 mm

Step-by-step explanation:

Information provided

$\bar X=2941$ represent the sample mean

$s=12$ represent the standard deviation

$n=25$ sample size      

$\mu_o =2944$ 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 test if the true mean for this case is equal to 2944 mm, the system of hypothesis would be:      

Null hypothesis:$\mu = 2944$      

Alternative hypothesis:$\mu \neq 2944$      

The statistic for this case would be given by:

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

Replacing the info given we got:

$t=\frac{2941-2944}{\frac{12}{\sqrt{25}}}=-1.250$      

The degrees of freedom are given by:

$df=n-1=25-1=24$  

The p value would be given by:

$p_v =2*P(t_{24}<-1.250)=0.223$  

Since the p value is higher than 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2944 mm