Answer:
The data provide sufficient evidence to indicate a difference in the mean stopping distances of the two types of automobiles
Step-by-step explanation:
Formulate the null and alternate hypothesis as
H0: u1= u2 against the claim Ha: u1≠ u2
The data given is
y`1=118, y`2=109 s21=102, s22=87
The test statistic used is t= y1`- y2`/ Sp√1/n1+ 1/n2
n1= n2= 64
And
Sp² = (n1-1) s12 + (n2-1) s22 / n1+n2-2
Sp² = 63*102+6*87/64+64-2
Sp² = 94.5
Sp= 9.7211
Putting the values in the test statistic
t= 118-108/9.7211√1/64+1/64
t= 5.237
Now finding the value of t at ∝=0.01 for two tailed test
t( 126,0.005)= ±2.6154
At alpha= 0.01 the null hypothesis is rejected.
Now finding the value of t at ∝=0.05 for two tailed test
t( 126,0.025)= ±1.979
Even at alpha= 0.05 the null hypothesis is rejected.
Also at alpha= 0.05 the null hypothesis is rejected because the critical value is ±1.657
Hence it is concluded that the data provide sufficient evidence to indicate a difference in the mean stopping distances of the two types of automobiles