Question

g To do this, he selects 25 bags of this brand at random and determines the net weight of each. He finds the sample mean to be 13.99 and the sample standard deviation to be s = 0.24. Calculate the test statistic.

Answer 1

Answer:

Null hypothesis:$\mu = 14$  

Alternative hypothesis:$\mu \neq 14$

$t=\frac{13.99-14}{\frac{0.24}{\sqrt{25}}}=-0.208$    

Step-by-step explanation:

Assuming this previous info: Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. Net weights actually vary slightly from bag to bag and are normally distributed with mean . A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is different with advertised and so intends to test the hypotheses

Data given and notation  

$\bar X=13.99$ represent the mean height for the sample  

$s=0.24$ represent the sample standard deviation for the sample  

$n=25$ sample size  

$\mu_o =14$ represent the value that we want to test

$\alpha$ 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 true mean is equal to 14 or no, the system of hypothesis would be:  

Null hypothesis:$\mu = 14$  

Alternative hypothesis:$\mu \neq 14$  

If we analyze the size for the sample is < 30 and we don't 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:  

$t=\frac{13.99-14}{\frac{0.24}{\sqrt{25}}}=-0.208$