Data on earnings were published for a particular country - Leaving Cert Mathematics - Question 9 - 2016

To find the income level at which the government will stop paying the subsidy:

1. Identify the z-score that corresponds to the lower 10% of a normal distribution, which is roughly -1.28.
2. Use the z-score formula to find the income level:
   $x = \mu + z \cdot \sigma$
   where $z = -1.28$, $\mu = 39400$, and $\sigma = 12920$.
   $x = 39400 + (-1.28) \cdot 12920 \approx 22862.40$
   Rounding to the nearest euro, the income level is €22,862.

1. State the null hypothesis ($H_0$) and the alternative hypothesis ($H_1$):
   • $H_0: \mu = 39400$
   • $H_1: \mu \neq 39400$
2. Calculate the z-score for the sample mean of €38,280:
   $z = \frac{38280 - 39400}{\frac{12920}{\sqrt{1000}}} \approx -2.74$
3. Determine the significance:
   • Compare the calculated z-score with critical z-values for a 5% significance level (approximately -1.96).
   • Since -2.74 < -1.96, reject the null hypothesis.
     Thus, there is significant evidence to conclude that the mean income has changed.

To create a 95% confidence interval:

1. Compute the confidence interval using the formula:
   $\bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}$
   where $\bar{x} = 26974$, $z \approx 1.96$, $\sigma = 5120$, and $n = 400$.
2. Calculate the margin of error:
   $ME = 1.96 \cdot \frac{5120}{\sqrt{400}} = 1.96 \cdot 256 = 502.4$
3. Thus, the confidence interval is:
   $[26974 - 502.4, 26974 + 502.4] \approx [26471.6, 27476.4]$
   Rounding gives us a confidence interval of €26,472 to €27,476.

One reason the research institute might create a sampling distribution of means from large random samples of farm size is to apply the Central Limit Theorem. This theorem states that regardless of the original distribution of the data, the distribution of sample means will tend to be normally distributed if the sample size is sufficiently large. This allows for more reliable statistical inference.

To find the sample size (n):

1. The margin of error (E) is given by the formula:
   $E = z \cdot \frac{\sigma}{\sqrt{n}}$
   where $E = 0.045$, and for a 95% confidence level, $z \approx 1.96$.
2. Rearranging the formula for n gives:
   $n = \left( \frac{z \cdot \sigma}{E} \right)^2$
   Assuming $\sigma$ is the standard deviation for farmers
   We can calculate the appropriate value once $\sigma$ is known.