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 ft

A. State the null and alternate hypothesis

B. Compute the test statistic

C. State long jupers conclusion (you can use p value or Critical value)

Answer 1

Solution:

Claim:  jump distance is less than 16 feet i.e   < 16

n = 19

= 13.2

= 1.5

Use = 10% = 0.10

A) Hypothesis are

H0 : = 16 or 16 (null hypo.)

H1 :   < 16 (alternative)

B)The test statistic z is given by

z =

= (13.2 - 16) / (1.5/19)

= -8.14

C) Now , observe that ,there is < sign in H1. So , the test is left tailed.

p value = P(Z < z)

= P(Z < -8.14)

= 0.000 (use z table)

Since p value is less than , we reject the null hypothesis and conclude that her jump distance is less than 16 feet