Another way to demonstrate the effectiveness of this “Calories Counter” program

Question

Another way to demonstrate the effectiveness of this “Calories Counter” program was to examine the heaviest of its members. Sixteen members were randomly selected and originally, their mean weight was 185 pounds. After two months on the program, their mean weight was 179 pounds with a standard deviation of 10 pounds. With such a large standard deviation and a relatively small aggregate decrease in weight, the volunteers and the public were still skeptical as to whether the diet plan actually worked. So you, as the statistician, step in and analyze the data. So, from these data, test the claim at 99% confidence that the diet actually worked. a) What are the Null and Alternative Hypotheses? (1) b) Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1) c) Conduct the test (i.e., determine the test value) (2) d) State the Decision and Interpretation (1, 1) e) What is the Pvalue? (2) f) Using this same information, prepare the 99% confidence interval for this diet plan. (2) g) Does the 99% confidence interval contain the before average weight? Discuss the information / decision made in parts i) and k). (1,1)

Explanation / Answer

HYPOTHESIS

Given that,
population mean(u)=185
sample mean, x =179
standard deviation, s =10
number (n)=16
null, Ho: =185
alternate, their mean weight was gradually decreased, H1: <185
level of significance, = 0.01
from standard normal table,left tailed t /2 =2.602
since our test is left-tailed
reject Ho, if to < -2.602
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =179-185/(10/sqrt(16))
to =-2.4
| to | =2.4
critical value
the value of |t | with n-1 = 15 d.f is 2.602
we got |to| =2.4 & | t | =2.602
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -2.4 ) = 0.01491
hence value of p0.01 < 0.01491,here we do not reject Ho
ANSWERS
---------------
null, Ho: =185
alternate, H1: <185
test statistic: -2.4
critical value: -2.602
decision: do not reject Ho
p-value: 0.01491

CONFIDENCE INTERVAL
DIRECT METHOD
given that,
sample mean, x =179
standard deviation, s =10
sample size, n =16
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 15 d.f is 2.95
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 = [ 179 ± t a/2 ( 10/ Sqrt ( 16) ]
= [ 179-(2.95 * 2.5) , 179+(2.95 * 2.5) ]
= [ 171.625 , 186.375 ]

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