Question

Name of volunteers	score on federation of planets
jaynes	218
carleton	224
spock	208
picard	222
ryker	200
la forge	219
armstrong	198
stryker	204
carpenter	210
grisson	189
glen	201
sheppard	184

The following is a list of The Federation of Planets’ Qualifying Scores for a randomly selected group of 12 of the volunteers who were selected for the 2-sample tests. Assume that the population of the pool, from which the volunteers were selected, is normally distributed.

The mean qualifying scores for all members of the Federation, who are qualified as crew members, is 209 with a standard deviation of 12.2. Is there reason to suspect that the scores of the sample of volunteers above is less than the mean score of all Federation members at the .05 level of significance? Use the p-value approach and the Classical approach. Use calculator, excel, or textbook tables as needed.

1. Set up the null and alternative hypothesis statements:

2. Find the t-score
3. Show all steps, (whether from Excel, StatCrunch, or TI-84 +) used.
4. Write out your conclusion of the test. Is there sufficient statistical evidence to suspect that the scores of the sample of volunteers is less than the scores of the Federation pool of crew members? Explain by discussing the Type I Error and what this means in terms of the hypothesis test.

Answer 1

Given that,
population mean(u)=209
sample mean, x =206.4167
standard deviation, s =12.8379
number (n)=12
null, Ho: μ=209
alternate, H1: μ<209
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.796
since our test is left-tailed
reject Ho, if to < -1.796
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =206.4167-209/(12.8379/sqrt(12))
to =-0.697
| to | =0.697
critical value
the value of |t α| with n-1 = 11 d.f is 1.796
we got |to| =0.697 & | t α | =1.796
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.6971 ) = 0.25012
hence value of p0.05 < 0.25012,here we do not reject Ho
ANSWERS
---------------
i.
null, Ho: μ=209
alternate, H1: μ<209
ii.
test statistic: -0.697
critical value: -1.796
iii.
decision: do not reject Ho
p-value: 0.25012
iv.
we do not have enough evidence to support the claim that the scores of the sample of volunteers above is less than the mean score of all Federation members