Question

Based on a random sample of 25 units of product X, the average weight is 102 lb and the sample standard deviation is 10 lb. We would like to decide whether there is enough evidence to establish that the average weight for the population of product X is greater than 100 lb. Assume the population is normally distributed. Using the critical value rule, at α = .01. What is the appropriate conclusion at 99% confident level?

Answer 1

Answer:

$t=\frac{102-100}{\frac{10}{\sqrt{25}}}=1$      

$t_{crit}= 2.49$

Since the calculated value is lower than the critical value we have enough evidence to FAIL to reject the null hypothesis

So then we can conclude that the population mean is not significantly higher than 100

Step-by-step explanation:

Data given and notation      

$\bar X=102$ represent the mean score for the sample  

$s=10$ represent the standard deviation for the sample      

$n=25$ sample size      

$\mu_o =100$ represent the value that we want to test    

$\alpha=0.01$ represent the significance level for the hypothesis test.    

z 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 is higher than 100, the system of hypothesis would be:    

Null hypothesis:$\mu \leq 100$      

Alternative hypothesis:$\mu > 100$      

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:      

     

Calculate the critical value

We need to find the degrees of freedom first, given by:

$df= n-1= 25-1 = 24$

Since we are conducting a right tailed test we need to find a critical value on the t distribution with 24 degrees of freedom that accumulates 0.01 of the area on the right and we got:

$t_{crit}= 2.49$

Since the calculated value is lower than the critical value w have enough evidence to FAIL to reject the null hypothesis

So then we can conclude that the population mean is not significantly higher than 100