Question

Let 𝑥₁, 𝑥₂, … , 𝑥₁₀ be the observed values of a random sample of size 10 from an 𝑁(𝜃, 𝜎 2 ) distribution, where 𝜃 ∈ ℝ and 𝜎 > 0 are unknown parameters. If
𝑥̅=\(\frac{1}{10}\) \(∑^{10}_{i=1}x_i=0\) and s=\(\sqrt{\frac{1}{9}∑^{10}_{i=1}(x_i-\overline{x})^2=2,}\)
then based on the values of 𝑥̅ and 𝑠 and using Student’s 𝑡-distribution with 9 degrees of freedom, 90% confidence interval(s) for 𝜃 is/are

• A. (−0.8746, ∞)
• B. (−0.8746, 0.8746)
• C. −1.1587, 1.1587)
• D. (−∞, 0.8746)

Answer 1

The Correct Option is A, C, D

To determine the 90% confidence interval for the mean \(\theta\) based on a sample from a normal distribution with unknown mean and variance, we shall use the t-distribution. Here, the sample mean \(\overline{x} = 0\), and the sample standard deviation \(s = 2\). The degrees of freedom for our sample size of 10 is \(n-1 = 9\).

The formula for the confidence interval is given by:

\(CI = \left( \overline{x} - t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}, \overline{x} + t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \right)\) 

where \(t_{\alpha/2}\) is the t-value for a 90% confidence interval with 9 degrees of freedom. For 90% confidence, each tail of the distribution covers 5%, hence we look for the t-value corresponding to 0.05 (this value can be looked up in the t-distribution table or calculated using statistical software). For 9 degrees of freedom, this is approximately 1.833.

Substituting the values:

• Sample mean \(\overline{x} = 0\)
• Standard deviation \(s = 2\)
• Sample size \(n = 10\)
• t-value \(t_{\alpha/2} = 1.833\)

Calculate the interval:

\(CI = \left( 0 - 1.833 \cdot \frac{2}{\sqrt{10}}, 0 + 1.833 \cdot \frac{2}{\sqrt{10}} \right)\)

Simplify:

\(CI = \left( 0 - \frac{3.666}{\sqrt{10}}, 0 + \frac{3.666}{\sqrt{10}} \right)\)

Approximating further:

\(CI = \left( 0 - 1.1587, 0 + 1.1587 \right)\)

Therefore, the 90% confidence interval for \(\theta\) is \((-1.1587, 1.1587)\).

Choice analysis:

• The correct answer \((-1.1587, 1.1587)\) corresponds with our calculations.
• The other intervals given \((-0.8746, \infty)\) and \((-\infty, 0.8746)\) do not align with the calculations based on the t-distribution with 9 degrees of freedom, which indicates symmetry about the sample mean.

Thus, the correct confidence interval is (-1.1587, 1.1587).