The percentage of farmers using fertilizers in Tanzania was known to be 35%. The drought and other events of the last few years are believed to have had a potential impact on the proportion of farmers usingfertilizers.Aninternationalaidprogramwantstotestwhetherornotthepercentageisstillaround 35% and gathers a random sample of 150 farmers. The findings reveal that 68 of the farmers use fertilizers. Conduct the test at 5% level of significance and state your conclusion
To conduct the hypothesis test, we need to set up the null and alternative hypotheses.
Null hypothesis (H0): The percentage of farmers using fertilizers in Tanzania is still around 35%.
Alternative hypothesis (H1): The percentage of farmers using fertilizers in Tanzania is different from 35%.
The test statistic used for this hypothesis test is the proportion test. We will calculate the test statistic using the formula:
z = (p̂ - p) / √((p * (1 - p)) / n)
where p̂ is the sample proportion, p is the hypothesized proportion (35% or 0.35), and n is the sample size.
p̂ = 68/150 = 0.453
z = (0.453 - 0.35) / √((0.35 * (1 - 0.35)) / 150)
= 0.103 / √(0.2275 / 150)
= 0.103 / 0.033883
≈ 3.042
Now, we need to find the critical value for a two-tailed test at a 5% level of significance. Since the test is two-tailed, we divide the significance level by 2, resulting in a critical value of 1.96.
Since the test statistic (3.042) is greater than the critical value (1.96), we reject the null hypothesis.
Conclusion: The findings provide sufficient evidence to suggest that the percentage of farmers using fertilizers in Tanzania is different from 35%.