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

Question

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

μ

B -

μ

A = 0 and alternative hypothesis as

μ

B -

μ

A

\neq

0

Then I did my calculations which were as following:

Test Statistic = (B -A ) - (

μ

B -

μ

A) / 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

{\bar{x}}_{1} - {\bar{x}}_{2}

is approximately normal with mean

μ_{1} - μ_{2}

and standard deviation

\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{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} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{{.62}^{3} * 3 + {.94}^{2} * 5}{8}}

giving a standard error of

{\bar{x}}_{1} - {\bar{x}}_{2}

of

s_{p} \sqrt{\frac{1}{n_{1}} + \frac{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