Samples are taken from two different types of honey and the viscosity is measured. Honey A:

Osmarq5ltp 2022-05-09 Answered
Samples are taken from two different types of honey and the viscosity is measured.

Honey A:
Mean: 114.44
S.D : 0.62
Sample Size: 4

Honey B:
Mean: 114.93
S.D: 0.94
Sample Size: 6

Assuming normal distribution, test at 5% significance level whether there is a difference in the viscosity of the two types of honey?

Here's what I did:

I took my null hypothesis as μB - μA = 0 and alternative hypothesis as μB - μA 0

Then I did my calculations which were as following:

Test Statistic = (B -A ) - ( μB - μA) / sqrt {(variance B / sample size B) + (variance A / sample size A)}

This gave me test statistic as = 0.49/0.49332 that is equal to 0.993

However the test statistic in the book solution is given as 0.91. What am I doing wrong?
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Answers (1)

candydulce168nlid
Answered 2022-05-10 Author has 11 answers
The book is using a pooled standard error, one of two possible methods. The central limit theorem says that x ¯ 1 x ¯ 2 is approximately normal with mean μ 1 μ 2 and standard deviation σ 1 2 n 1 + σ 2 2 n 2 , but when the population sizes are small and it can be reasonably assumed the populations have the same variance, then a pooled standard error is used with s p = ( n 1 1 ) s 1 2 + ( n 2 1 ) s 2 2 n 1 + n 2 2 = .62 3 3 + .94 2 5 8 giving a standard error of x ¯ 1 x ¯ 2 of s p 1 n 1 + 1 n 2 . This is to combine more data into the calculation of the standard error of each population mean, which are assumed to be equal for these purposes.

So the t-distribution with n 1 + n 2 2 = 8 degrees of freedom has test statistic
t = ( x ¯ 1 x ¯ 2 ) ( μ 1 μ 2 ) s p 1 4 + 1 6 = 0.9096458
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