Question

Two Sample t-test (16pts): Suppose you are interested in deciding if the 1990 Toyota Four Runner has been equally reliable as the 1990 Honda Passport. You go out a randomly sample of 5 people who own a 1990 Toyota and 5 other people who own a 1990 Honda and you ask them how often they have to take their vehicles in for maintenance. Here are your data (in thousands of miles): Toyota: 31 35 32 34 30 Honda: 29 33 28 31 27

a) State the null and alternative hypotheses (2pts)

b) Compute the means and standard deviations of the two samples (2-pts)

c) Compute the two sample t-statistic (2 pts) c) How many degrees of freedom do you have? (3pts)

d) Compute the P-value (4pts)

e) At an alpha = 0.05 would you accept or reject the null hypothesis? (3pts)

Answer 1

n1 = 5

n2=5

s1 =2.0736

s2 =2.4083

claim : 1990 Toyota Four Runner has been equally reliable as the 1990 Honda Passport

a) Null and alternative hypothesis is

Vs

Level of significance = 0.05

b)

sample mean for Toyota =

standard deviations for  Toyota = s1 = 2.0736

sample mean for  Honda =

standard deviations for Honda = s2 =2.4083

c)

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is

Vs

Test statistic is

F = Larger variance / Smaller variance = 5.7999 / 4.2998 = 1.3489

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 5 - 1 = 4

Degrees of freedom for denominator = n2 - 1 = 5 - 1 = 4

Critical value = 6.388     ( using f-table )

F test statistic < critical value ,we fail to reject null hypothesis.

Conclusion: Population variances are equal.

So we have to use pooled variance.

Formula

d.f = n1 + n2 – 2 = 5 + 5 - 2 = 8

d)

p-value = 0.0843 ( using t table )

e) At

p-value , Failed to Reject Ho

conclusion : There is a insufficient evidence to conclude that 1990 Toyota Four Runner has been equally reliable as the 1990 Honda Passport