Question

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project. Dylan randomly selects 32 student athletes and 45 students who do not play any sports to walk along a board that was 16 feet long and 2 inches wide and raised 6 inches off the ground. Dylan records the number of times each participant touched the ground. The sample of athletes had a mean of 3.7 touches with a standard deviation of 1.1. The sample of non-athletes had a mean of 4.1 touches with a standard deviation of 1.3. Let μ1 be the population mean number of touches for student athletes, and let μ2 be the population mean number of students who do not play any sports. Dylan is testing the alternative hypothesis Ha:μ1−μ2<0 and assumes that the population standard deviation of the two groups of students are equal. If the p-value is greater than 0.05 and less than 0.10 and the significance level is α=0.01, what conclusion could be made about the balance of student athletes and the balance of students who do not play sports? Identify all of the appropriate conclusions to the hypothesis test below.

Answer 1

Answer:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

$p_v =P(t_{75}$  

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

Step-by-step explanation:

Data given and notation

$\bar X_{A}=3.7$ represent the mean for athletes  

$\bar X_{NA}=4.1$ represent the mean for non athletes  

$s_{A}=1.1$ represent the sample standard deviation for athletes  

$s_{NA}=1.3$ represent the sample standard deviation for non athletes

$n_{A}=32$ sample size for the group 2  

$n_{NA}=45$ sample size for the group 2  

$\alpha=0.01$ 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 population mean for athletes is lower than the population mean for non athletes, the system of  hypothesis would be:  

Null hypothesis:$\mu_{A}-\mu_{NA}\geq 0$  

Alternative hypothesis:$\mu_{A} - \mu_{NA}< 0$  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

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

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:  

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

P value  

We need to find first the degrees of freedom given by:

$df=n_A +n_{NA}-2=32+45-2=75$

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

$p_v =P(t_{75}$  

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.