A well respected wild life researcher recently conducted a study of wild Jackalopes. He trapped and weighed 60 adult
wild Jackalopes in an attempt to gauge their health. As a
population, if a species loses weight it could be a signal
that the health of the population is in danger. Research conducted 10 years prior found that the population mean was 69.9 pounds. Test the claim that population mean of wild Jackalopes is equal to 69.9 lbs. Use a significance level of .01.

Claim: _
Null Hypothesis: _
Alternative Hypothesis: _
Test Statistic: (state the statistic and find it’s value):
Find the pvalue for the given test statistic:
At the 0.01 significance level would you reject the null or fail to reject the null? Justify your answer.
What final conclusion could you draw from your research?

Claim : men weigh of wild jackalopes is 69.9

The null hypothesis : H0 : μ = 69.9

Alternative hypothesis : H1 : μ ≠ 69.9

Test statistic = −2.447085

P value = 0.0174

Conclusion :

Fail to Reject the Null hypothesis

Step-by-step explanation:

From the question given :

The claim is that : mean weight of wild jackalopes is still the same as 10 years with a mean weight of 69.9 lbs.

The null hypothesis : H0 : μ = 69.9

Alternative hypothesis : H1 : μ ≠ 69.9

Using calculator :

Sample mean (x) = 66

Sample standard deviation (s) = 12.345

The test statistic t :

(x - μ) / (s/√n)

n = sample size = 60

(66 - 69.9) / (12.345 / √60)

t = −2.447085

P value at α 0.01, df = 59 is 0.0174

Since the p value is > 0.01, the result is not significant at 0.01. Therefore, we fail to reject the Null

step-by-step explanation:

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6