Question

Dr. Krauze wants to see how cell phone use impacts reaction time. To test this, Dr. Krauze conducted a study where participants are randomly assigned to one of two conditions while driving: a cell phone or no cell phone. Participants were then instructed to complete a driving simulator course where reaction times (in milliseconds) were recorded by how quickly they hit the breaks in response to a dog running in the middle of the road during the course. Below are the data. What can Dr. Krauze conclude with an α = 0.01?

cell phone	no cell phone
235 250 239 243 232	232 238 227 228 227

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- no cell phone reaction time cell phone simulator course dog running
Condition 2:
---Select--- no cell phone reaction time cell phone simulator course dog running

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =
Decision:  ---Select--- Reject H0 Fail to reject H0

Answer 1

Answers: a) The appropriate test statistic is Independent samples t-test because the participants here are chosen randomly and to the conditions of using cellphones or not using cellphones.

We are to test whether the average reaction times in both groups are statistically different from one another.

b) Question not properly given

c) We construct the null and alternative hypotheses as H0: mu1 = mu2 vs Ha: mu1 not equal to mu2 where mu1 and mu2 are the unknown population means. We have the test statistic as T = (x1bar-x2bar)/sqrt((s1*s1/n1)+(s2*s2/n2)), where x1bar, x2bar are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes.

We reject H0 iff |T(observed)| > t(alpha/2,v) where t(Alpha/2,v) is the upper alpha/2 point of a Student's t-distribution with v degrees of freedom. v= (((s1/n1)+(s2/n2))^2) / (((s1/n1)^2)/(n1-1))+(((s2/n2)^2)/(n2-1)).

Here, the value of the test statistic is T(observed) =  2.4771 and the critical value is = 3.501534. Thus we see that |T(observed)| < t(alpha/2,v). Hence, we fail to reject H0 and conclude on the basis of the given sample at a 1% level of significance that there is no statistically significant difference between the two groups.

[The answer is obtained using R-software. The code and output are attached below.]

R-code:

#x1= cell phones x2 = no cell phones
x1=c(235,250,239,243,232)
x2=c(232,238,227,228,227)
x1bar=mean(x1)
x2bar=mean(x2)
n1=length(x1)
n2=length(x2)
s1=var(x1)
s2=var(x2)
t=(x1bar-x2bar)/sqrt((s1/n1)+(s2/n2))
t
vn=((s1/n1)+(s2/n2))^2
vd=(((s1/n1)^2)/(n1-1))+(((s2/n2)^2)/(n2-1))
v=vn/vd
qt(0.995,v)

Output:

R Console > #x1= cell phones x2 = no cell phones >xl=c(235,250,239,243,232) >> X2=c(232, 238, 227, 228, 227) > Xlbar=mean (xl) X2bar=mean (x2) > nl-length (x2) > n2=length (x2) > Sl=var (xl) 52=var (X2) > t= (xlbar-x2bar) /sqrt((31/nl) + (s2/n2) ) [1] 2.477118 > vn=((sl/nl) + (32/n2))^2 >> vd=(((sl/nl) 2) / (nl-l))+(((s2/02)*2) 7 (n2-1)) > y= vv >qt(0.995, V) [1] 3.501534