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I was solving this question in my book and I realized the answer at the back and my answer doesn't match. I wanted to verify whether my process is correct or not?

The question is as follows:

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? Help!

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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 $\mu_1-\mu_2$ 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 $s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\sqrt{\frac{.62^3*3+.94^2*5}{8}}$ 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 $n_1+n_2-2=8$ degrees of freedom has test statistic $$t^*=\frac{(\bar x_1-\bar x_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac 1 4+\frac 1 6}}=0.9096458$$

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  • $\begingroup$ thank you so much $\endgroup$ – min Feb 16 at 19:22

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