Aleasing firm claims that the mean number of miles driven annually, u , in its leased cars is less than 12800 miles. a random sample of 50 cars leased from this firm had a mean of 12499 annual miles driven. it is known that the population standard deviation of the number of miles driven in cars from this firm is 3140 miles. is there support for the firm's claim at the 0.05 level of significance?

for 0.05 level of significance p value is greater than so it cant not reject

Step-by-step explanation:

given data

mean = 12499 miles

sample n = 50

standard deviation = 3140 miles

μ = 12800 miles

to find out

Is there support for claim 0.05 level significance

solution

we find here first standard statistic test for z

z =  ( mean - μ)  / ( SD × √n )    1

put here value

z = ( 12499 - 12800)  / ( 3140 / √50 )

z = -0.67783

and we can find from table

P value for z is -0.67783 is 0.2489

so for 0.05 level of significance p value is greater than so it cant not reject

step-by-step explanation:

given expression is .

now we need to simplify the given expression .

we know that square root is equivalent to the power 1/2. so that means we can take half of the exponents to get the square root of given expression.

expanding that will give a very very large number so let's keep answer in exponent form.