Answer:
We accept null hypothesis
Step-by-step explanation:
We assume a normal distribution
The population mean μ₀ = 65 mph
Sample mean μ = 63,2 mph ( calculated from data )
Sample standard deviation σ = 7,309
Sample size n = 8
Degree of freedom is n - 1 8 - 1 = 7
As n < 30 we have to use the t-student test
We will do our test with a confidence interval of 95 % that means α = 5 %
or α = 0,05 and as we are going through a two-tail test α/2 = 0,025
Test Hypothesis:
Null Hypothesis: H₀ μ = μ₀
Alternate Hypothesis Hₐ μ ≠ μ₀
From t-student table for the degree of freedom 7, α/2 = 0,025 two-tail test we find tc
tc = 2,365
And calculate ts as
ts = ( μ - μ₀ ) / σ /√n
ts = ( 63,2 - 65 ) / 7,309/ √8
ts = - 1,8 *2,828/ 7,309
ts = - 5,091 /7,309
ts = - 06965
Now we compare ts and tc
tc = 2,365 or tc = - 2,365 ( by simmetry) tc = -2,37
and ts = -0,06965 ts = - 0,07
As |ts| < |tc|
ts is in the acceptance zone so we accept null hypothesis
