Question

Data were obtained from a randomly selected sample of 25 UMass students. The study observed a mean weight of 177 lbs (standard deviation of 16 lbs). Assume the mean weight of all US college students is 160 lbs. What statistical test should be used to assess if the UMass students mean weight differs from the US students mean weight? Select one:
a. 1-sample t-test
b. 1-sample z-test
c. ANOVA
d. F test

Answer 1

Answer:

Since the sample size is <30 and we don't know the population standard deviation the correct test is:

a. 1-sample t-test

Step-by-step explanation:

1) Data given and notation  

$\bar X=177$ represent the mean weigth for the sample

$s=16$ represent the sample standard deviation

$n=25$ sample size  

$\mu_o =160$ 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)  

Since the sample size is <30 and we don't know the population standard deviation the correct test is:

a. 1-sample t-test

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 160 lbs, the system of hypothesis would be:  

Null hypothesis:$\mu = 160$  

Alternative hypothesis:$\mu \neq 160$  

If we analyze the size for the sample is < 30 and 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:  

$t=\frac{177-160}{\frac{16}{\sqrt{25}}}=5.3125$    

P-value

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

$df=n-1=25-1=24$  

Since is a two sided test the p value would be:  

$p_v =2*P(t_{(24)}>5.3125)=1.88x10^{-5}$  

Conclusion  

If we compare the p value and the significance level assumed for example $\alpha=0.01$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the mean is different from 160 at 1% of signficance.