Question

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from two different age categories are collected, and the following summary data on stance duration is presented. It is assumed that both age categories have equal population (true) variances.
Age Category Sample Size Sample Mean Sample Standard Deviation
Age Category 1 13 810 67
Age Category 2 19 770 56

Test the hypothesis that the mean of age category 1 is greater than the mean of age category 2 at 5% significance level.
Calculate P-Value.

Answer 1

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$  

$df=n_1 +n_2 -2=13+19-2=30$  

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

$p_v =P(t_{30}>1.771)=0.0434$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1  

$\bar X_{2}=770$ represent the mean for sample 2  

$s_{1}=67$ represent the sample standard deviation for 1  

$s_{2}=56$ represent the sample standard deviation for 2  

$n_{1}=13$ sample size for the group 2  

$n_{2}=19$ sample size for the group 2  

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}\leq 0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}> 0$  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$  

P value  

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

$p_v =P(t_{30}>1.771)=0.0434$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2