Suppose a long jumper claims that her jump distance is less than 16 feet, on average. several of her teammates do not believe her, so the long jumper decides to do a hypothesis test, at a 10% significance level, to persuade them. she makes 19 jumpes. the mean distance of the sample jumps is 13.2 feet. the long jumper knows from experience that the standard deviation of her jump distance is 1.5 fta. state the null and alternate hypothesisb. compute the test statisticc. state long jupers conclusion (you can use p value or critical value)

Claim : Jump distance is less than 16 feet.

A )

n = 19

Since n <30 we will use t test

The mean distance of the sample jumps is 13.2 feet. i.e. x = 13.2 feet

The standard deviation of her jump distance is 1.5 ft i.e. s = 1.5

Formula :

Substitute the values:

B)

degree of freedom df = n-1 = 19-1 = 18

t critical > t statistics

So, we accept the null hypothesis

C) So, the claim is wrong that ump distance is less than 16 feet.

Hence long jump distance must be greater than or equal to 16

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

true

step-by-step explanation: