Question

1. When people go on a diet they should lost on average 8 pounds in the first month of their diet. We let 68 try the “statistician’s diet” (every time you get hungry you decide to do some statistics instead) and find the average weight loss for this group of 68 people was 12 pounds with a standard deviation of 10 pounds for the first month they were on the diet.

a) At the .01 level did the diet increase the amount of weight lost? (p = 0)

b) Interpret the p-value obtained

c) Interpret the 99% confidence interval (computer says “over 14.2”)

d) Based on the results of this test would you go on this statistician’s diet? Explain

e) What is the only type of error that could have occurred

Answer 1

1.

a.
Given that,
population mean(u)=8
sample mean, x =12
standard deviation, s =10
number (n)=68
null, Ho: μ=8
alternate, H1: μ>8
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.383
since our test is right-tailed
reject Ho, if to > 2.383
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =12-8/(10/sqrt(68))
to =3.2985
| to | =3.2985
critical value
the value of |t α| with n-1 = 67 d.f is 2.383
we got |to| =3.2985 & | t α | =2.383
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 3.2985 ) = 0.00078
hence value of p0.01 > 0.00078,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=8
alternate, H1: μ>8
test statistic: 3.2985
critical value: 2.383
decision: reject Ho
b.
p-value: 0.00078
we have enough evidence to support the claim that diet increase the amount of weight lost.
c.
TRADITIONAL METHOD
given that,
sample mean, x =12
standard deviation, s =10
sample size, n =68
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 10/ sqrt ( 68) )
= 1.213
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 67 d.f is 2.651
margin of error = 2.651 * 1.213
= 3.215
III.
CI = x ± margin of error
confidence interval = [ 12 ± 3.215 ]
= [ 8.785 , 15.215 ]
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DIRECT METHOD
given that,
sample mean, x =12
standard deviation, s =10
sample size, n =68
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 67 d.f is 2.651
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 12 ± t a/2 ( 10/ Sqrt ( 68) ]
= [ 12-(2.651 * 1.213) , 12+(2.651 * 1.213) ]
= [ 8.785 , 15.215 ]
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interpretations:
1) we are 99% sure that the interval [ 8.785 , 15.215 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
d.
yes,
based on the test results the statistician diet go up because it reject the null hyppothesis and interval value increases( range between 8.785 , 15.215).
e.
based on test results,
type 1 error is possible because it reject the null hypothesis.