Respuesta :
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
Answer:
-0.70
Step-by-step explanation:
For the tabulated value the mean is calculated as:
Mean = (60.5 + 63.2 + 54.7 + 51.6 + 72.3 + 70.7 + 67.2 + 65.4)/8
= 505.6/8
Mean \bar{x}= 63.2
and population mean as assumption u= 65
and given that the sample standard deviation is: s= 7.309
The test statistic is calculated as:
Ζ = Τ –μ 63.2 - 65 = -0.696 -0.70 S
Hence the T statistic would be -0.70
