Question

5(a)What is the probability that out of 200 pieces of randomly selected glass, more than fifty-five of them are defective. [5 marks]

A sample of 12 of bags of Calbie Chips were weighed (to the nearest gram), and listed here as follows.

219, 226, 217, 224, 223, 216, 221, 228, 215, 229, 225, 229 Find a 95% confidence interval for the mean mass of bags of Calbie Chips.

[9 marks]

(b) Professor GeniusAtCalculus has two lecture sections (A and B) of the same 4th year Advanced Calculus (AMA 4301) course in Semester 2. She wants to investigate whether section A students maybe ”smarter” than section B students by comparing their perfor- mances in the midterm test. A random sample of 12 students were taken from section A, with mean midterm test score of 78.8 and standard deviation 8.5; and a random sample of 9 students were taken from section B, with mean midterm test score of 86 and standard deviation 9.3. Assume the population standard deviations of midterm test scores for both sections are the same. Construct the 90% confidence interval for the difference in midterm test scores of the two sections. Based on the sample midterm test scores from the two sections, can Professor GeniusAtCalculus conclude that there is any evidence that one section of students are ”smarter” than the other section? Justify your conclusions.

[8 marks]

(c) The COVID-19 (coronavirus) mortality rate of a country is defined as the ratio of the number of deaths due to COVID-19 divided by the number of (confirmed) cases of COVID-19 in that country. Suppose we want to investigate if there is any difference between the COVID-19 mortality rate in the US and the UK. On April 18, 2020, out of a sample of 671,493 cases of COVID-19 in the US, there was 33,288 deaths; and out of a sample of 109,754 cases of COVID-19 in the UK, there was 14,606 deaths. What is the 92% confidence interval in the true difference in the mortality rates between the two countries? What can you conclude about the difference in the mortality rates between the US and the UK? Justify your conclusions. [8 marks]

Answer 1

5.

a.
TRADITIONAL METHOD
given that,
sample mean, x =222.666
standard deviation, s =4.818
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.818/ sqrt ( 12) )
= 1.391
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 * 1.391
= 3.061
III.
CI = x ± margin of error
confidence interval = [ 222.666 ± 3.061 ]
= [ 219.605 , 225.727 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =222.666
standard deviation, s =4.818
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 = [ 222.666 ± t a/2 ( 4.818/ Sqrt ( 12) ]
= [ 222.666-(2.201 * 1.391) , 222.666+(2.201 * 1.391) ]
= [ 219.605 , 225.727 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 219.605 , 225.727 ] 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.
TRADITIONAL METHOD
given that,
mean(x)=78.8
standard deviation , s.d1=8.5
number(n1)=12
y(mean)=86
standard deviation, s.d2 =9.3
number(n2)=9
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (11*72.25 + 8*86.49) / (21- 2 )
s^2 = 78.246
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 78.246 * (1/12+1/9) )
=3.901
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 19 d.f is 1.729
margin of error = 1.729 * 3.901
= 6.744
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (78.8-86) ± 6.744 ]
= [-13.944 , -0.456]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=78.8
standard deviation , s.d1=8.5
sample size, n1=12
y(mean)=86
standard deviation, s.d2 =9.3
sample size,n2 =9
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 78.8-86) ± t a/2 * sqrt( 78.246 * (1/12+1/9) ]
= [ (-7.2) ± 6.744 ]
= [-13.944 , -0.456]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-13.944 , -0.456]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
c.
i.
TRADITIONAL METHOD
given that,
sample one, x1 =33288, n1 =671493, p1= x1/n1=0.0496
sample two, x2 =14606, n2 =109754, p2= x2/n2=0.1331
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.0496*0.9504/671493) +(0.1331 * 0.8669/109754))
=0.0011
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.08
from standard normal table, two tailed z α/2 =1.75
margin of error = 1.75 * 0.0011
=0.0019
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.0496-0.1331) ±0.0019]
= [ -0.0854 , -0.0817]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =33288, n1 =671493, p1= x1/n1=0.0496
sample two, x2 =14606, n2 =109754, p2= x2/n2=0.1331
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.0496-0.1331) ± 1.75 * 0.0011]
= [ -0.0854 , -0.0817 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 92% sure that the interval [ -0.0854 , -0.0817] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 92% of these intervals will contains the difference between
true population mean P1-P2
ii.
Given that,
sample one, x1 =33288, n1 =671493, p1= x1/n1=0.05
sample two, x2 =14606, n2 =109754, p2= x2/n2=0.133
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.08
from standard normal table, two tailed z α/2 =1.751
since our test is two-tailed
reject Ho, if zo < -1.751 OR if zo > 1.751
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.05-0.133)/sqrt((0.061*0.939(1/671493+1/109754))
zo =-106.917
| zo | =106.917
critical value
the value of |z α| at los 0.08% is 1.751
we got |zo| =106.917 & | z α | =1.751
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -106.9171 ) = 0
hence value of p0.08 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -106.917
critical value: -1.751 , 1.751
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference in the mortality rates between the US and the UK