Question

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 guage their health. As a population, if a species loses significant 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 the population mean of wild Jackalopes is still 69.9 pounds.

76.4	52.9	78.4	62.8	89.2
75.8	72.1	64.8	57.9	52.3
59.0	59.5	78.5	72.8	52.7
69.1	58.2	78.0	43.6	54.0
75.8	72.0	75.0	84.2	63.2
55.8	72.9	68.0	63.3	76.9
59.6	62.7	38.6	47.9	52.8
42.7	55.4	68.2	65.2	33.5
59.1	65.4	62.1	69.3	75.3
78.2	76.9	87.6	71.5	63.4
84.2	48.9	78.4	79.0	64.5
88.8	76.3	59.9	62.3	65.0

1. Using your graphing calculator, put your data into your L1 list. Double check your data to make sure it is correct. Go to Stat, Calc, 1-VarStats. Determine the mean and standard deviation for your sample:

Sample Mean: 65.963 lbs

Sample Standard Deviation: 12.429 lbs

2. What is the initial claim of this problem?

3. For the original claim, what is the null and alternative hypothesis?

• A.

  Ho:μ=69.9
  HA:μ<69.9

• B.

  Ho:μ=69.9
  HA:μ>69.9

• C.

  Ho:μ>69.9
  HA:μ=69.9

• D.

  Ho:μ<69.9
  HA:μ=69.9

• E. H₀: μ = 69.9
  H_A: μ ≠ 69.9

Answer 1

2) The claim of this problem is that the population mean of wild Jackalopes is still 69.9 pounds.

3) The null and alternative hypotheses are

69.9 Ho : μ HA : μ69.9

The test statistic is

X 69.9 t = S 65.963 69.9 t = 12.429 V60 t = -2,4536062

The P-value is

p-value 20T(-2.4536062) p-value 0.017

So we reject the null hypothesis at 5% significance since .

p-value 0.017 <a= 0.05