Answer :
Solution :
Let [tex]p_1[/tex] and [tex]p_2[/tex] represents the proportions of the seeds which germinate among the seeds planted in the soil containing [tex]3\%[/tex] and [tex]5\%[/tex] mushroom compost by weight respectively.
To test the null hypothesis [tex]H_0: p_1=p_2[/tex] against the alternate hypothesis [tex]H_1:p_1 \neq p_2[/tex] .
Let [tex]\hat p_1, \hat p_2[/tex] denotes the respective sample proportions and the [tex]n_1, n_2[/tex] represents the sample size respectively.
[tex]$\hat p_1 = \frac{74}{155} = 0.477419[/tex]
[tex]n_1=155[/tex]
[tex]$p_2=\frac{86}{155}=0.554839[/tex]
[tex]n_2=155[/tex]
The test statistic can be written as :
[tex]$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}[/tex]
which under [tex]H_0[/tex] follows the standard normal distribution.
We reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance, if the P-value [tex]<0.05[/tex] or if [tex]|z_{obs}|>Z_{0.025}[/tex]
Now, the value of the test statistics = -1.368928
The critical value = [tex]\pm 1.959964[/tex]
P-value = [tex]$P(|z|> z_{obs})= 2 \times P(z< -1.367928)$[/tex]
[tex]$=2 \times 0.085667$[/tex]
= 0.171335
Since the p-value > 0.05 and [tex]$|z_{obs}| \ngtr z_{critical} = 1.959964$[/tex], so we fail to reject [tex]H_0[/tex] at [tex]0.05[/tex] 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 [tex]\text{proportion}[/tex] of the seeds that [tex]\text{germinate differs}[/tex] with the percent of the [tex]\text{mushroom compost}[/tex] in the soil.