Question

Statistics!! Please help, 10 points and brainliest!

Answer 1

Answer:

$t=\frac{70.1-65.2}{\frac{2.52}{\sqrt{30}}}=10.65$    

$p_v =P(t_{(29)}>10.65)=7.76x10^{-12}$  

And the best conclusion for this case would be:

D. The p-value is less than 0.00001. There is sufficient data to reject the null hypothesis.

Step-by-step explanation:

Data given and notation  

$\bar X=70.1$ represent the sample mean

$\sigma=2.52$ represent the population standard deviation  

$n=30$ sample size  

$\mu_o =65.2$ 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 mean is higher than 65.2, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 65.2$  

Alternative hypothesis:$\mu > 65.2$  

If we analyze the size for the sample is = 30 but 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:  

$t=\frac{70.1-65.2}{\frac{2.52}{\sqrt{30}}}=10.65$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=30-1=29$  

Since is a one side right tailed test the p value would be:  

$p_v =P(t_{(29)}>10.65)=7.76x10^{-12}$  

And the best conclusion for this case would be:

D. The p-value is less than 0.00001. There is sufficient data to reject the null hypothesis.