Question

A collegiate long jumper is hoping to improve his distance with improved conditioning. His new conditioning routine should help him be able to jump further than he has in the past. Before the conditioning began, the jumper was averaging 24.5 feet. After his new routine was finished, he jumps a series of 30 jumps over the course of a week. His new average is 25.2 feet with a standard deviation of 2.47 feet. Does this new sample give evidence that he has actually increased his average jumping distance?

1. What is the parameter for this hypothesis test? (choose one)? ?μ ?¯y¯ ?p ?̂ p^

2. What are the null and alternative hypotheses? a) ?0H0:___ b) ??HA:___

3. This test is (choose one): One-sided Two-sided

4. Calculate the test statistic. Round to two decimal places

5. Your test statistic is needs to be compared to a t distribution with n-1 degrees of freedom. Draw a line at that test statistic value (t) on this t distribution with 29 degrees of freedom, and shade the area corresponding for the p-value for the hypothesis test. To calculate the p-value, you need to use Stat → Calculators → T in StatCrunch. Make sure you type in 29 for degrees of freedom.

6. What is the p-value for this hypothesis test?

7. Compare your p-value to the significance level α=0.05. What is your decision based on the p-value? (Choose one letter.) A. Since the p-value <α<α, reject the null hypothesis (?0H0). B. Since the p-value <α<α, do not reject the null hypothesis (?0H0). C. Since the p-value>α>α, reject the null hypothesis (?0H0). D. Since the p-value >α>α, do not reject the null hypothesis (?0H0).

8. Based on your decision, what is the conclusion? (Choose one letter.) A. At the 0.05 level of significance, we do have sufficient evidence to say that the athlete has actually increased his average jumping distance. B. At the 0.05 level of significance, we do not have sufficient evidence to say that the athlete has actually increased his average jumping distance.

9. Suppose your alternative hypothesis had been ??:?≠24.5 What would have been the p-value for that hypothesis test?

Answer 1

Given that,
population mean(u)=24.5
sample mean, x =25.2
standard deviation, s =2.47
number (n)=30
null, Ho: μ=24.5
alternate, H1: μ>24.5
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.699
since our test is right-tailed
reject Ho, if to > 1.699
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25.2-24.5/(2.47/sqrt(30))
to =1.5523
| to | =1.5523
critical value
the value of |t α| with n-1 = 29 d.f is 1.699
we got |to| =1.5523 & | t α | =1.699
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.5523 ) = 0.06572
hence value of p0.05 < 0.06572,here we do not reject Ho
ANSWERS
---------------
1.
parameter of interest
population mean(u)=24.5
sample mean, x =25.2
standard deviation, s =2.47
number (n)=30
2.
null, Ho: μ=24.5
alternate, H1: μ>24.5
3.
one sided tail
4.
test statistic: 1.5523
5.
critical value: 1.699
decision: do not reject Ho
6.
p-value: 0.06572
7.
p value is greater than alpha value
8.
we do not have enough evidence to support the claim that new sample give evidence that he has actually increased his average jumping distance
9.
Given that,
population mean(u)=24.5
sample mean, x =25.2
standard deviation, s =2.47
number (n)=30
null, Ho: μ=24.5
alternate, H1: μ!=24.5
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25.2-24.5/(2.47/sqrt(30))
to =1.5523
| to | =1.5523
critical value
the value of |t α| with n-1 = 29 d.f is 2.045
we got |to| =1.5523 & | t α | =2.045
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.5523 ) = 0.1314
hence value of p0.05 < 0.1314,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=24.5
alternate, H1: μ!=24.5
test statistic: 1.5523
critical value: -2.045 , 2.045
decision: do not reject Ho
p-value: 0.1314
we do not have enough evidence to support the claim that new sample give evidence that he has average is 24.5