Question

In: Statistics and Probability

Find solutions to the following problems. Use the appropriate t-test to test for significant mean differences...

Find solutions to the following problems. Use the appropriate t-test to test for significant mean differences in the following research scenarios. Report all relevant information, including hypotheses, degrees of freedom, critical value, and a statement about significance.

A researcher wants to see if antihistamines will increase the amount of time in seconds it takes participants to react to a surprise stimulus. He first collects the participants' reaction times while not on antihistamines, and then gives them the dose of antihistamine. One hour later, he collects the participants' reaction times again. Participant ID Reaction Time Before Antihistamine Reaction Time After Antihistamine

Participant ID Reaction Time
Before Antihistamine
Reaction Time
After Antihistamine
1 1.37 1.45
2 2.45 2.90
3 1.95 2.01
4 3.01 3.25
5 .95 1.07
6 1.50 1.71
7 2.05 2.07
8 2.31 2.30
9 3.07 3.15
10 1.99 2.11
x 2.065 2.202
s 0.68 0.72

Solutions

Expert Solution

Before Antihistamine After Antihistamine Difference
1.37 1.45 -0.08
2.45 2.9 -0.45
1.95 2.01 -0.06
3.01 3.25 -0.24
0.95 1.07 -0.12
1.5 1.71 -0.21
2.05 2.07 -0.02
2.31 2.3 0.01
3.07 3.15 -0.08
1.99 2.11 -0.12

Sample mean of the difference using excel AVERAGE(), x̅d = -0.1370

Sample standard deviation of the difference usinf excel STDEV.S(), sd =0.1343

Sample size, n =10

Null and Alternative hypothesis: incease the amount of time means Before < After

Ho : µd = 0    

H1 : µd < 0    

Test statistic:     

t = (x̅d)/(sd/√n) = -3.2270    

df = n-1 = 9    

Critical value :  

Let α = 0.05

Left tailed critical value, t_c = T.INV(0.05,9) = -1.833

p-value :     

Left tailed p-value = T.DIST(-3.2270,9,1 ) = 0.0052

Decision:     

p-value < α, Reject the null hypothesis.

There is enough evidence to claim that antihistamines will increase the amount of time in seconds it takes participants to react to a surprise stimulus at 0.05 significance level.


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