Question

The weights of soy patties sold by a diner are normally distributed. A random sample of 25 patties yields a mean weight of 4.2 ounces with a sample standard deviation of 0.5 ounces. At the 0.05 level of​ significance, perform a hypothesis test to see if the true mean weight is less than 4 ounces. What is the correct calculated value of the test​ statistic?

Answer 1

Answer:

$t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2$

The degrees of freedom are given by:

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

And the p value would be given by:

$p_v = P(t_{24}>2) =0.0285$

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate      

Step-by-step explanation:

Data provided

$\bar X=4.2$ represent the sample mean for the weigths

$s=0.5$ represent the sample standard deviation

$n=25$ sample size      

$\mu_o =4$ represent the value that we want to analyze

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

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to conduct a hypothesis in order to check if the true mean weigth is less than 4 or not, the system of hypothesis would be:      

Null hypothesis:$\mu \leq 4$      

Alternative hypothesis:$\mu > 4$      

The statistic for this case is given by:

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

Replacing the info given we got:

$t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2$  

The degrees of freedom are given by:

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

And the p value would be given by:

$p_v = P(t_{24}>2) =0.0285$

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate