Question

In an experiment to learn whether Substance M can help restore memory, the brains of 20 rats were treated to damage their memories. First, the rats were trained to run a maze. After a day, 10 rats (determined at random) were given M and 7 of them succeeded in the maze. Only 2 of the 10 control rats were successful. The two-sample z test for "no difference" against "a significantly higher proportion of the M group succeeds"
a) should not be used because the Large Counts condition is violated.
b) gives z = 2.25, P < 0.02
c) gives z = 2.60, P < 0.005
d) should not be used because the Random condition is violated
e) gives z = 2.25, P < 0.04 but not < 0.02

Answer 1

The number of rats in the sample are less than 30, therefore, according

to the central limit theorem, cannot approximate the population.

Response:

a) Should not be used because the Large Counts Condition is violated

Which condition are necessary for the hypothesis test?

The proportion of the rats given M that succeeded, $\hat p_1$ = 7 ÷ 10 = 0.7

The proportion of the control rats that succeeded, $\hat p_2$ = 2 ÷ 10 = 0.2

H₀: $\hat p_1$ = $\hat p_2$

Hₐ: $\hat p_1$ > $\hat p_2$

The z-test formula for the difference between two proportion is given as follows;

$Z= \mathbf{\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p})\left (\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right )}}}$

Where;

$\hat p = \dfrac{7 + 2}{10 + 10} = 0.45$

$Z=\dfrac{0.7-0.2}{\sqrt{0.45\times (1-0.45) \times \left (\dfrac{1}{10}+\dfrac{1}{10} \right )}} \approx \mathbf{ 2.25}$

p = 1 - p(z < 2.25) = 1 - 0.9878 = 0.0122

However, the number of rats in the sample, n are less than 30

Therefore;

The sample is not large enough for the test compared to the population,

and for a normal approximation which gives;

a) Should not be used because the Large Counts Condition is violated

Learn more about normal distribution here:

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