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.

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

-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

a

step-by-step explanation: