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

To find the income level for the lowest 10% of income earners, we need to determine the z-score that corresponds to the 10th percentile:

From z-tables, we find:

$z_{10\%} \approx -1.28$

Now we convert the z-score back to the raw score using the mean and standard deviation:

$x = μ + zσ = 39400 + (-1.28)(12920) \approx 22862.40 \Rightarrow €22,862$.

Let:

• Null Hypothesis (H₀): The mean income of full-time workers has not changed, i.e., H₀: μ = 39400
• Alternative Hypothesis (H₁): The mean income has changed, i.e., H₁: μ ≠ 39400

We calculate the test statistic:

Sample mean (x̄) = €38,280, n = 1000

$z = \frac{x̄ - μ}{\sigma/\sqrt{n}} = \frac{38280 - 39400}{12920/ ext{√}1000} \approx -2.74$

At the 5% significance level, we refer to the z-table: The critical z-values are approximately -1.96 and +1.96. Since -2.74 < -1.96, we reject the null hypothesis. This suggests there is significant evidence that the mean income has changed since the national data were published.

To create a 95% confidence interval for the mean income of full-time farmers, we use the sample mean (x̄) and standard deviation (σ) to compute the interval:

Mean income: €26,974 Standard deviation: €5,120 Sample size: n = 400

a) Calculate the margin of error:

$E = z_{0.025} \times \frac{σ}{\sqrt{n}} = 1.96 \times \frac{5120}{\sqrt{400}} = 1.96 \times 256 = 502.86$

b) The confidence interval is:

$26,974 - 502.86 \leq μ \leq 26,974 + 502.86$

Thus, the interval is:

$[26,471.14, 27,476.86]$.

One reason for creating a sampling distribution of means from large random samples is to ensure a normal distribution of data, which allows for valid statistical inference. The Central Limit Theorem states that the means of large enough samples will approximate a normal distribution, regardless of the original data's distribution. This enables accurate estimation of parameters and hypothesis testing.

To find the sample size (n) required for a given margin of error (E) at a certain confidence level:

Given:

• Margin of error (E) = 4.5% = 0.045
• We use the formula for margin of error:

$E = z \times \frac{σ}{\sqrt{n}}$

Assuming a 95% confidence level, z ≈ 1.96:

$0.045 = 1.96 \times \frac{1}{\sqrt{n}}$

Rearranging gives:

$\sqrt{n} = \frac{1.96}{0.045} \implies n = \left(\frac{1.96}{0.045}\right)^2 \approx 493.827$

Thus, rounding to the nearest whole number, we find n ≈ 494.