## Answers

1.

Given that,

mean(x)=132.86

standard deviation , s.d1=15.34

number(n1)=10

y(mean)=127.44

standard deviation, s.d2 =18.23

number(n2)=21

null, Ho: u1 = u2

alternate, H1: u1 != u2

level of significance, α = 0.05

from standard normal table, two tailed t α/2 =2.045

since our test is two-tailed

reject Ho, if to < -2.045 OR if to > 2.045

calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)

s^2 = (9*235.316 + 20*332.333) / (31- 2 )

s^2 = 302.224

we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))

to=132.86-127.44/sqrt((302.224( 1 /10+ 1/21 ))

to=5.42/6.679

to=0.811

| to | =0.811

critical value

the value of |t α| with (n1+n2-2) i.e 29 d.f is 2.045

we got |to| = 0.811 & | t α | = 2.045

make decision

hence value of |to | < | t α | and here we do not reject Ho

p-value: two tailed ( double the one tail ) - ha : ( p != 0.8115 ) = 0.4235

hence value of p0.05 < 0.4235,here we do not reject Ho

ANSWERS

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null, Ho: u1 = u2

alternate, H1: u1 != u2

test statistic: 0.811

critical value: -2.045 , 2.045

decision: do not reject Ho

p-value: 0.4235

we do not have enough evidence to support the claim that underlying mean difference in blood pressure between the two groups.

2.

TRADITIONAL METHOD

given that,

mean(x)=132.86

standard deviation , s.d1=15.34

number(n1)=10

y(mean)=127.44

standard deviation, s.d2 =18.23

number(n2)=21

I.

calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)

s^2 = (9*235.316 + 20*332.333) / (31- 2 )

s^2 = 302.224

II.

standard error = sqrt(S^2(1/n1+1/n2))

=sqrt( 302.224 * (1/10+1/21) )

=6.679

III.

margin of error = t a/2 * (standard error)

where,

t a/2 = t -table value

level of significance, α = 0.05

from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 29 d.f is 2.045

margin of error = 2.045 * 6.679

= 13.659

IV.

CI = (x1-x2) ± margin of error

confidence interval = [ (132.86-127.44) ± 13.659 ]

= [-8.239 , 19.079]

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DIRECT METHOD

given that,

mean(x)=132.86

standard deviation , s.d1=15.34

sample size, n1=10

y(mean)=127.44

standard deviation, s.d2 =18.23

sample size,n2 =21

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 = [( 132.86-127.44) ± t a/2 * sqrt( 302.224 * (1/10+1/21) ]

= [ (5.42) ± 13.659 ]

= [-8.239 , 19.079]

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interpretations:

1. we are 95% sure that the interval [-8.239 , 19.079]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