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

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 $\mu$B - $\mu$A = 0 and alternative hypothesis as $\mu$B - $\mu$A $\ne$ 0

Then I did my calculations which were as following:

Test Statistic = (B -A ) - ($\mu$B - $\mu$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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candydulce168nlid
The book is using a pooled standard error, one of two possible methods. The central limit theorem says that ${\overline{x}}_{1}-{\overline{x}}_{2}$ is approximately normal with mean ${\mu }_{1}-{\mu }_{2}$ and standard deviation $\sqrt{\frac{{\sigma }_{1}^{2}}{{n}_{1}}+\frac{{\sigma }_{2}^{2}}{{n}_{2}}}$, 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 ${s}_{p}=\sqrt{\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}}=\sqrt{\frac{{.62}^{3}\ast 3+{.94}^{2}\ast 5}{8}}$ giving a standard error of ${\overline{x}}_{1}-{\overline{x}}_{2}$ 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 ${n}_{1}+{n}_{2}-2=8$ degrees of freedom has test statistic
${t}^{\ast }=\frac{\left({\overline{x}}_{1}-{\overline{x}}_{2}\right)-\left({\mu }_{1}-{\mu }_{2}\right)}{{s}_{p}\sqrt{\frac{1}{4}+\frac{1}{6}}}=0.9096458$