40% . Accent3 Py Merge& Center . $ , % , 4% Conditional Format as 40%. Accent4 4

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40% . Accent3 Py Merge& Center . $ , % , 4% Conditional Format as 40%. Accent4 40% Accents her mat Painter Formatting Table Font Alignment Number Styles e two samples below are independent from Normal populations with equal variances. Sample 1: Sample 2: 2.700 490 35 346,000 45,800 12) Question 1A Can you prove 1 > 2 at a; 0.1? State the hypothesis in terms of 1-2. Step 1: H0 Ha: Step 2 tep 3 rill in row 35 if the hypothesis is one tailed Reject H0 if tsta I Fill in row 39·t the hypotheses is two tailed the smealer number mst be fypad first. 37 Reject HD if Step 4:

Explanation / Answer

1.a.
Given that,
mean(x)=2700
standard deviation , s.d1=588.217
number(n1)=35
y(mean)=490
standard deviation, s.d2 =214.0093
number(n2)=11
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.301
since our test is right-tailed
reject Ho, if to > 1.301
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (34*345999.239089 + 10*45799.980486) / (46- 2 )
s^2 = 277772.134861
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=2700-490/sqrt((277772.134861( 1 /35+ 1/11 ))
to=2210/182.176725
to=12.131078
| to | =12.131078
critical value
the value of |t | with (n1+n2-2) i.e 44 d.f is 1.301
we got |to| = 12.131078 & | t | = 1.301
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 12.1311 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 =u2
alternate, H1: u1 > u2
test statistic: 12.131078
critical value: 1.301
decision: reject Ho
p-value: 0
we have enough evidence to support the claim

1.b

TRADITIONAL METHOD
given that,
mean(x)=2700
standard deviation , s.d1=588.217
number(n1)=35
y(mean)=490
standard deviation, s.d2 =214.0093
number(n2)=11
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (34*345999.239 + 10*45799.98) / (46- 2 )
s^2 = 277772.135
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 277772.135 * (1/35+1/11) )
=182.177
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.1
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 44 d.f is 1.68
margin of error = 1.68 * 182.177
= 306.057
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (2700-490) ± 306.057 ]
= [1903.943 , 2516.057]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=2700
standard deviation , s.d1=588.217
sample size, n1=35
y(mean)=490
standard deviation, s.d2 =214.0093
sample size,n2 =11
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 = [( 2700-490) ± t a/2 * sqrt( 277772.135 * (1/35+1/11) ]
= [ (2210) ± 306.057 ]
= [1903.943 , 2516.057]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [1903.943 , 2516.057]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

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