Question

A random sample of 225 measurements is selected from a population, and the sample mean and standard deviation are x =32.5 and s = 30.0, respectively. It is claimed that the population mean exceeds 30. State the null and an appropriate alternative hypothesis, and perform a test at 5% significance level.

Answer 1

Answer:

$t=\frac{32.5-30}{\frac{30}{\sqrt{225}}}=1.25$    

$p_v =P(t_{(224)}>1.25)=0.106$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is actually its significantly higher than 30.

Step-by-step explanation:

Data given and notation  

$\bar X=32.5$ represent the sample mean  

$s=30$ represent the sample standard deviation

$n=225$ sample size  

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ 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 exceeds 30, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 30$  

Alternative hypothesis:$\mu > 30$  

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:  

$t=\frac{32.5-30}{\frac{30}{\sqrt{225}}}=1.25$    

P-value

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

$df=n-1=225-1=224$  

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

$p_v =P(t_{(224)}>1.25)=0.106$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is actually its significantly higher than 30.