It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. suppose that the sample mean for a random sample of 100 cars is 27.2 miles and assume the standard deviation is 2.4 miles. now suppose the car producer wants to test the hypothesis that μ, the mean number of miles per gallon, is 25.9 against the alternative hypothesis that it is not 25.9. conduct a test using α=.05 by giving the following:

Step-by-step explanation:

Given that the sample mean for a random sample of 100 cars is 27.2 miles and assume the standard deviation is 2.4 miles.

H0: x bar = 25.9

Ha: x bar ≠ 25.9

(Two tailed test at 95% confidence)

Alpha =0.05

The hypothetical mean is 25.900  

The actual mean is 27.200  

The difference between these two values is 1.300  

The   95% confidence interval of this difference:  

From -3.462 to 6.062

t = 0.5417  

 df = 99  

 standard error of difference = 2.400

The two-tailed P value equals 0.5893  

 Since p >0.05 this difference is considered to be not statistically significant.  

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

-232

step-by-step explanation: