Question

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from soil. Out of 155 seeds planted in soil containing 3% mushroom compost by weight, 74 germinated. Out of 155 seeds planted in soil containing 5% mushroom compost by weight, 86 germinated. Can you conclude that the proportion of seeds that germinate differs with the percent of mushroom compost in the soil

Answer 1

Let
`p_1` and
`p_2` represents the proportions of the seeds which germinate among the seeds planted in the soil containing
`3\%` and
`5\%` mushroom compost by weight respectively.

To test the null hypothesis
`H_0: p_1=p_2` against the alternate hypothesis
`H_1:p_1 \\eq p_2` .

Let
`\hat p_1, \hat p_2` denotes the respective sample proportions and the
`n_1, n_2` represents the sample size respectively.

`$\hat p_1 = (74)/(155) = 0.477419`

`n_1=155`

`$p_2=(86)/(155)=0.554839`

`n_2=155`

The test statistic can be written as :

`$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{(\hat p_1 * (1-\hat p_1))/(n_1)} + (\hat p_2 * (1-\hat p_2))/(n_2)}}`

which under
`H_0` follows the standard normal distribution.

We reject
`H_0` at
`0.05` level of significance, if the P-value
`<0.05` or if
`|z_(obs)|>Z_(0.025)`

Now, the value of the test statistics = -1.368928

The critical value =
`\pm 1.959964`

P-value =
`$P(|z|> z_(obs))= 2 * P(z< -1.367928)$`

`$=2 * 0.085667$`

= 0.171335

Since the p-value > 0.05 and
`$|z_(obs)| \\gtr z_(critical) = 1.959964$`, so we fail to reject
`H_0` at
`0.05` level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the
`\text{proportion}` of the seeds that
`\text{germinate differs}` with the percent of the
`\text{mushroom compost}` in the soil.