Let's solve each part of the problem step by step.

1.2.1 Compute the point estimate of the average score of all candidates who wrote the test

The point estimate for the average score is simply the sample mean ($\bar{x}$).

Given the dataLet's solve each part of the problem step by step.

1.2.1 Compute the point estimate of the average score of all candidates who wrote the test

The point estimate for the average score is simply the sample mean ($\bar{x}$).

Given the
65, 70, 78, 65, 50, 99, 55, 76, 58, 81, 88, 70, 63, 77, 50, 65, 55, 78, 91, 95, 56, 54, 70, 76, 54, 70, 86, 84

To calculate the sample mean: $\bar{x} = \frac{\text{Sum of all scores}}{\text{Number of scores}}$

I'll compute the sum and then divide by the number of observations (28).

$\text{Sum of scores} = 65 + 70 + 78 + \dots + 86 + 84 = 1864$ $\bar{x} = \frac{1864}{28} \approx 66.57$

So, the point estimate of the average score of all candidates is 66.57.

1.2.2 Construct a 95% confidence interval estimate of the true population mean of candidates' scores in the numeracy test

To construct a 95% confidence interval for the population mean, we'll use the following formula: $\text{Confidence Interval} = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean (66.57)
• $s$ is the sample standard deviation
• $n$ is the sample size (28)
• $t_{\alpha/2}$ is the t-score for the desired confidence level, with degrees of freedom $n-1 = 27$

Step 1: Calculate the sample standard deviation ($s$):

The sample standard deviation is calculated using: $s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}$

I'll calculate the standard deviation based on the given data.

Step 2: Find the $t$-score:

For a 95% confidence level and 27 degrees of freedom, the $t$-score ($t_{0.025}$) can be found using a t-table or calculator. It is approximately $t_{0.025} \approx 2.052$.

Step 3: Calculate the confidence interval:

Substituting the values: $\text{Margin of Error} = 2.052 \times \frac{s}{\sqrt{28}}$

Let's calculate these values.

First, we find the standard deviation ($s$):

$s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{27}} \approx 13.56$

Now, compute the margin of error: $\text{Margin of Error} = 2.052 \times \frac{13.56}{\sqrt{28}} \approx 5.26$

Finally, the confidence interval is: $\text{Confidence Interval} = 66.57 \pm 5.26$ $= (61.31, 71.83)$

So, the 95% confidence interval estimate of the true population mean of candidates' scores is (61.31, 71.83).

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Would you like further details on these calculations or have any questions?

5 Relative Questions:

1. How does the sample size affect the width of a confidence interval?
2. What is the difference between a t-distribution and a normal distribution?
3. Why is the t-distribution used instead of the normal distribution when constructing confidence intervals?
4. How would the confidence interval change if the sample standard deviation were larger?
5. What does the confidence interval tell us about the population mean?

Tip: Remember that a larger sample size generally leads to a narrower confidence interval, giving a more precise estimate of the population mean.