Question

 

1. Claim: most adults do not have hypertension. When 983 randomly selected adults were tested, it was found that 70.9% of them do not have hypertension.

               a.) express the original claim in symbolic form

               b.) identify the null and alternative hypothesis

               c.) find the z statistic

2. One of Mendel’s famous genetics experiments yielded 580 peas, with 428 of them green and 152 yellow.

               a.) find a 95% confidence interval estimate of the percentage of green peas.

               b.) based on his theory, Mendel expected that 75% of the offspring peas would be green. Given the percentage of offspring green peas is not 75%, do the results contradict his theory? Why or why not?

3. Generic aspirin tablets are supposed to contain 325 mg of aspirin. Below are the measured amounts found in randomly selected tablets.

               330        358        318        338        317        329        339        324        409        248        357               315

               a.) construct a 95% confidence interval estimate of the mean amount of aspirin in tablets (keep in mind which statistic (t vs z) that you should use…).

               b.) do the samples appear to be acceptable to capture the amount of aspirin the tablets have claimed to contain?

4. Using the data regarding aspirin in question 3, what is the 95% CI of the variance?

              

Answer 1

1.
Given that,
possibile chances (x)=696.947
sample size(n)=983
success rate ( p )= x/n = 0.709
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.709-0.5/(sqrt(0.25)/983)
zo =13.105
| zo | =13.105
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =13.105 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 13.10548 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.5
alternate, H1: p!=0.5
b.
test statistic: 13.105
critical value: -1.96 , 1.96
c.
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that most adults do not have hypertension.

2.
a.
TRADITIONAL METHOD
given that,
possible chances (x)=428
sample size(n)=580
success rate ( p )= x/n = 0.7
I.
sample proportion = 0.7
standard error = Sqrt ( (0.7*0.3) /580) )
= 0
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table,FALSE z α/2 =1.645
margin of error = 1.645 * 0
= 0
III.
CI = [ p ± margin of error ]
confidence interval = [0.7 ± 0]
= [ 0.7 , 0.8]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=428
sample size(n)=580
success rate ( p )= x/n = 0.7
CI = confidence interval
confidence interval = [ 0.7 ± 1.645 * Sqrt ( (0.7*0.3) /580) ) ]
= [0.7 - 1.645 * Sqrt ( (0.7*0.3) /580) , 0.7 + 1.645 * Sqrt ( (0.7*0.3) /580) ]
= [0.7 , 0.8]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.7 , 0.8] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

b.
We are 95% sure that the interval [ 0.7 , 0.8]
given that based on his theory, Mendel expected that 75% of the offspring peas would be green
the percentage of offspring green peas is not 75%, here the range in 70% to 80%.it is in the range only.

3.
a.
TRADITIONAL METHOD
given that,
sample mean, x =331.833
standard deviation, s =37.121
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 37.121/ sqrt ( 12) )
= 10.716
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
margin of error = 2.201 * 10.716
= 23.586
III.
CI = x ± margin of error
confidence interval = [ 331.833 ± 23.586 ]
= [ 308.247 , 355.419 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =331.833
standard deviation, s =37.121
sample size, n =12
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
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 = [ 331.833 ± t a/2 ( 37.121/ Sqrt ( 12) ]
= [ 331.833-(2.201 * 10.716) , 331.833+(2.201 * 10.716) ]
= [ 308.247 , 355.419 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 308.247 , 355.419 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

b.
yes,
the samples appear to be acceptable to capture the amount of aspirin the tablets have claimed to contain
given value is Generic aspirin tablets are supposed to contain 325 mg of aspirin.
so that it is in the confidence interval range.

c.
CONFIDENCE INTERVAL FOR VARIANCE
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s^2 = variance
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since aplha =0.05
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.025 = 0.975
the two critical values ᴪ^2 left, ᴪ^2 right at 11 df are 21.92 , 3.8157
variance( s^2 )=968.516
sample size(n)=12
confidence interval = [ 11 * 968.516/21.92 < σ^2 < 11 * 968.516/3.8157 ]
= [ 10653.676/21.92 < σ^2 < 10653.676/3.816 ]
[ 486.025 , 2791.844 ]