Question

The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm and a sample standard deviation of 0.77 gm. Find the test statistic to see whether the candy bars are smaller than they are supposed to be.

Answer 1

Answer:

$t=\frac{55.82-56}{\frac{0.77}{\sqrt{49}}}=-1.636$      

Step-by-step explanation:

Data given and notation      

$\bar X=55.82$ represent the sample mean

$s=0.77$ represent the standard deviation for the sample      

$n=49$ sample size      

$\mu_o =56$ 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 determine if the mean is less than 56, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 56$      

Alternative hypothesis:$\mu < 56$      

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:      

$t=\frac{55.82-56}{\frac{0.77}{\sqrt{49}}}=-1.636$