Question

Adam tabulated the values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6, 72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309. Adam reads that the average speed that cars drive on the highway is 65 mph. The t-test statistic for a two-sided test would be __________. Answer choices are rounded to the hundredths place.

Answer 1

-0.70

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

Answer 2

We accept null hypothesis

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