Q1.
From data we get : n1 = 7, n2 = 7,
= 14.14,
= 11.57, s1 = 1.345, s2 = 1.618
Here population variances are equal .
Hence 90% confidence interval is given by,
{ (
-
) - E, (
-
) + E }
Where,
Where,
= 1.488
c = 0.90, df = n1+n2-2 = 7+7 -2 = 12 ,
Hence tc =
------using Excel formula "=t.inv.2t(0.10,12)"
Hence the margin of error is,
= 1.42
Hence the 90% confidence interval is,
{ ( 14.14 -11.57) -1.42 , ( 14.14 -11.57 ) + 1.42 }
( 1.15, 3.99 )
b) The hypothesis are
H0:
v/s H1:
The test statistic is,
= 1.814
The critical value is given by,
df = n1+n2 -2 = 7+7 -2 = 12 , = 0.10
Hence the critical value =
------using Excel formula "=t.inv.2t(0.10,12)"
So the calculated value of t > critical value of t .
Hence we reject the null hypothesis.
Conclusion :
There is difference between two population means.
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