In: Statistics and Probability
A paper investigated the driving behavior of teenagers by observing their vehicles as they left a high school parking lot and then again at a site approximately
1 |
2 |
mile from the school. Assume that it is reasonable to regard the teen drivers in this study as representative of the population of teen drivers.
Male Driver |
Female Driver |
1.4 | -0.2 |
1.2 | 0.5 |
0.9 | 1.1 |
2.1 | 0.7 |
0.7 | 1.1 |
1.3 | 1.2 |
3 | 0.1 |
1.3 | 0.9 |
0.6 | 0.5 |
2.1 | 0.5 |
(a) Use a .01 level of significance for any hypothesis tests. Data consistent with summary quantities appearing in the paper are given in the table. The measurements represent the difference between the observed vehicle speed and the posted speed limit (in miles per hour) for a sample of male teenage drivers and a sample of female teenage drivers. (Use μmales − μfemales. Round your test statistic to two decimal places. Round your degrees of freedom down to the nearest whole number. Round your p-value to three decimal places.)
t | = | |
df | = | |
P | = |
(b) Do these data provide convincing support for the claim that, on
average, male teenage drivers exceed the speed limit by more than
do female teenage drivers?
YesNo
Working
We find the mean, variance and standard deviation of the data using Excel functions average, var.s and stdev.s respectively.
n | Mean (M) | Variance (SS) | Standard Deviation (SD) | |
Males | n1 = 10 | M1 = 1.46 | SS1 = 0.5493 | s1 = 0.7412 |
Females | n2 = 10 | M2 = 0.64 | SS2 = 0.2071 | s2 = 0.4551 |
The null and alternative hypotheses are
Ho : μmales = μfemales
Ha : μmales > μfemales
where μmales and μfemales are population
means for exceeding speed limits of males and females
respectively.
α = 0.01
We use independent Samples T test since population standard
deviation is unknown, and sample size is small
Using the given formulae, we get
Degrees of
Freedom
df1 = n1 - 1 , df2 = n2 - 1 , df = n1 + n2 - 2
df = 18
Pooled
Variance
Sp2 = 0.3782
Mean Squared Error
Sm1-m2
Sm1-m2 = 0.275
t-statistic
t-statistic = t = 2.9814
P-value
For t = 2.9814, df = 18 we find the one tailed p-value using t
tables or Excel function t.dist
p-value = 1 - t.dist(2.9814, 18, TRUE)
p-value = 0.004
Answer :
t | = | 2.98 |
df | = | 18 |
P | = | 0.004 |
b)
Decision
0.004 < 0.01
that is p-value is less than alpha.
Hence we Reject Ho.
Answer :
YES
The data provides convincing support at α = 0.01 to show that on
average, male teenage drivers exceed the speed limit by more than
do female teenage drivers.