Question

Consider the following hypothesis test: Cmu%20%5C%20%20%5Ctextless%20%5C%20%2040" id="TexFormula1" title="H_0 : \mu \geq 40 ; H_1 : \mu \ \textless \ 40" alt="H_0 : \mu \geq 40 ; H_1 : \mu \ \textless \ 40" align="absmiddle" class="latex-formula">A sample of 49 observations provides a sample mean of 38 and a sample standard deviation of 7. Compute the value of the test statistic.

Answer 1

Answer:

$t=\frac{38-40}{\frac{7}{\sqrt{49}}}=-2$    

Step-by-step explanation:

1) Data given and notation    

$\bar X=38$ represent the sample mean  

$s=7$ represent the population standard deviation for the sample    

$n=49$ sample size    

$\mu_o =40$ 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)

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the population mean is less than 40, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 1150$    

Alternative hypothesis:$\mu < 1150$    

We don't know the population deviation, so for this case we can use the 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".

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

$t=\frac{38-40}{\frac{7}{\sqrt{49}}}=-2$    

4) Calculate the P-value    

We need to calculate first the degrees of freedom, given by:

$df=n-1=49-1=48$

Since is a one-side lower test the p value would be:    

$p_v =P(t_{48}$

5) Conclusion    

If we compare the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can reject the null hypothesis, so there is enough evidence to conclude that the mean is less than 40.