Question

Cholesterol levels are a major concern to health insurance providers. Cholesterol levels in U.S. adults average 215 with a standard deviation of 30. If the cholesterol levels of a sample of 42 U.S. adults is taken, what is the probability that the mean cholesterol level of the sample 1) Will be less than 200? 2) Will be between 205 and 225? 3) Will be greater than 220?

2. Suppose that 36% of the clients of a company use smart phones. Draw a random sample of 200 clients, what is the probability that the sample proportion is 1) Less than 0.28? 2) Greater than 0.42?

3. A bottler wishes to ensure that the average of 16 ounces of fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 48 bottles. The mean weight of the juice in the sample is 15.8 ounces. Assume the population standard deviation is 0.8 ounce. Test his concern at the 5% significance level.

4. In a sample of 1022 U.S. adults, 73% say that they are not happy with the current investment climate. Is there evidence that the percentage has significantly decreased from the 74% reported previously? Do the test at the 1% level.

5. The percentage of bills being paid by Medicare has been 31%. An examination of 8368 recent bills reveals that 32% of these bills are being paid by Medicare. Is this evidence of a change in the percentage of bills being paid by Medicare? Do the test at the 10% level.

Answer 1

i.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 215
standard Deviation ( sd )= 30/ Sqrt ( 42 ) =4.6291
sample size (n) = 42
1.
the probability that the mean cholesterol level of the sample Will be less than 200
P(X < 200) = (200-215)/30/ Sqrt ( 42 )
= -15/4.6291= -3.2404
= P ( Z <-3.2404) From Standard NOrmal Table
= 0.0006
2.
the probability that the mean cholesterol level of the sample Will be between 205 and 225
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 205) = (205-215)/30/ Sqrt ( 42 )
= -10/4.6291
= -2.1602
= P ( Z <-2.1602) From Standard Normal Table
= 0.0154
P(X < 225) = (225-215)/30/ Sqrt ( 42 )
= 10/4.6291 = 2.1602
= P ( Z <2.1602) From Standard Normal Table
= 0.9846
P(205 < X < 225) = 0.9846-0.0154 = 0.9692
3.
the probability that the mean cholesterol level of the sample Will be greater than 220
P(X > 220) = (220-215)/30/ Sqrt ( 42 )
= 5/4.629= 1.0801
= P ( Z >1.0801) From Standard Normal Table
= 0.14
ii.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.36
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.36*0.64/200)
=0.0339
1.
the probability that the sample proportion is Less than 0.28
P(X < 0.28) = (0.28-0.36)/0.0339
= -0.08/0.0339= -2.3599
= P ( Z <-2.3599) From Standard Normal Table
= 0.0091
2.
the probability that the sample proportion is Greater than 0.42
P(X > 0.42) = (0.42-0.36)/0.0339
= 0.06/0.0339 = 1.7699
= P ( Z >1.77) From Standard Normal Table
= 0.0384
iii.
Given that,
population mean(u)=16
standard deviation, σ =0.8
sample mean, x =15.8
number (n)=48
null, Ho: μ=16
alternate, H1: μ!=16
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) = x-u/(s.d/sqrt(n))
zo = 15.8-16/(0.8/sqrt(48)
zo = -1.732
| zo | = 1.732
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.732 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.732 ) = 0.083
hence value of p0.05 < 0.083, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=16
alternate, H1: μ!=16
test statistic: -1.732
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.083
we do not have enough evidence to support the claim that A bottler wishes to ensure that the average of 16 ounces of fruit juice is used to fill each bottle.
iv.
Given that,
possibile chances (x)=746.06
sample size(n)=1022
success rate ( p )= x/n = 0.73
success probability,( po )=0.74
failure probability,( qo) = 0.26
null, Ho:p=0.74
alternate, H1: p<0.74
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.33
since our test is left-tailed
reject Ho, if zo < -2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.73-0.74/(sqrt(0.1924)/1022)
zo =-0.7288
| zo | =0.7288
critical value
the value of |z α| at los 0.01% is 2.33
we got |zo| =0.729 & | z α | =2.33
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.72882 ) = 0.23305
hence value of p0.01 < 0.23305,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.74
alternate, H1: p<0.74
test statistic: -0.7288
critical value: -2.33
decision: do not reject Ho
p-value: 0.23305
we do not have enough evidence to support the claim that the percentage has significantly decreased from the 74% reported previously
v.
Given that,
possibile chances (x)=2677.76
sample size(n)=8368
success rate ( p )= x/n = 0.32
success probability,( po )=0.31
failure probability,( qo) = 0.69
null, Ho:p=0.31
alternate, H1: p!=0.31
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.32-0.31/(sqrt(0.2139)/8368)
zo =1.9779
| zo | =1.9779
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =1.978 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.97791 ) = 0.04794
hence value of p0.1 > 0.0479,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.31
alternate, H1: p!=0.31
test statistic: 1.9779
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.04794
we have enough evidence to support the claim that change in the percentage of bills being paid by Medicare.