Question

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)

Answer 1

Answer:

Claim : Jump distance is less than 16 feet.

A ) $H_0:\mu \geq16\\H_a: \mu$

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 : $t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Substitute the values:

B) $t =\frac{13.2-16}{\frac{1.5}{\sqrt{19}}}$

$t =−8.136$

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

$\alpha = 0.1$

$t_{(\frac{\alpha}{2},df)}=1.73$

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