A leasing firm claims that the mean number of miles driven annually, , in its leased cars is less than miles. A random sample of cars leased from this firm had a mean of annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is miles. Assume that the population is normally distributed. Is there support for the firm's claim at the level of significance?
Perform a one-tailed test.
null hypothesis?
alternative hypothesis?
type of test statistic?
value of the test statistic?
the p-value?
Can we support the leasing firm's claim that the mean number of miles driven annually is less than 12800 miles?

Complete question is;

A leasing 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?

Perform a one-tailed test.

null hypothesis?

alternative hypothesis?

type of test statistic?

value of the test statistic?

the p-value?

Can we support the leasing firm’s claim that the mean number of miles driven annually is less than 12800 miles?

A) Null hypothesis; μ ≥ 12800

Alternative hypothesis; μ < 12800

B) type of test statistic: z-score test

C) test statistic; z = -0.678

D) p - value = 0.2489

E) No, we can't support the leasing firms claim

Step-by-step explanation:

We are given;

Population mean; μ = 12800 miles

Sample mean; x¯ = 12499 miles

Sample size; n = 50

Population standard deviation; σ = 3140

Let's define the hypothesis;

Null hypothesis; μ ≥ 12800

Alternative hypothesis; μ < 12800

The alternative hypothesis is the claim.

Since sample size is greater than 30, we will use z-score test.

Formula for z-score;

z = (x¯ - μ)/(σ/√n)

z = (12499 - 12800)/(3140/√50)

z = -0.678

From online p-value from z-score calculator attached, using z = -0.678 with one tailed hypothesis and significance value of 0.05,we have;

p - value = P(z < -0.678) = 0.2489

The p-value is greater than the significance value, so we will fail to reject the null hypothesis and conclude that there is no sufficient information to support the leasing firm's claim

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

the number of the people going.

step-by-step explanation:

the cost to buy one movie ticket is $7. if the total cost for the movie is a function of how many people go, then the input is : the number of the people going.