Aleasing firm claims that the mean number of miles driven annually, μ, in its leased cars is less than 13460 miles. a random sample of 50 cars leased from this firm had a mean of 12781 annual miles driven. it is known that the population standard deviation of the number of miles driven in cars from this firm is 1760 miles. is there support for the firm's claim at the 0.01 level of significance?

There is support for the firm's claim.

Step-by-step explanation:

We have to test the null hypotesis

The significance level is 0.01 (an area of 0.005 on both sides of the normal distribution curve). The interval that is defined by this significance levels are z1=-2.576 and z2=2.576.

We have to compute the probability value z

Since this value of z=-0.545 lies inside the interval [-2.576,2.576], we can not reject the null hypotesis. The value of the sample mean (M=12781) was not significantly different from the mean of the population the firm claims (μ=13460).

There is support for the firm's claim.

m a t h w a y. c o m

step-by-step explanation:

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕