Suppose you have observations \, X_1,X_2,X_3, X_4, X_5\, which are i.i.d. draws from a Gaussian distribution with unknown mean \mu and unknown variance \sigma ^2.
For all of the problems on this tab, suppose you are given the following:
\frac{1}{5} \sum _{i=1}^5 X_ i = 0.9, \qquad \frac{1}{5} \sum _{i=1}^5 X_ i^2 = 1.33
To test the null hypothesis H_0 : \mu = 0 versus the alternative hypothesis H_1 : \mu \neq 0 using the data above, which of the following test(s) is appropriate?
(Choose all that apply.)
t-test
Z-test: i.e. the test based on the central limit theorem
Wald's test
Compute the unbiased sample variance S.
(Enter a numerical answer accurate to at least 2 decimal places.)
S=\quad
Give an estimate the mean \mu and variance \sigma ^2. Use the maximum likelihood estimator.
\hat\mu = \quad
\hat\sigma ^2 =\quad
Find the value of the \text {t}-statistic for testing the hypotheses above:
H_0 : \mu = 0 \quad \text {versus} \quad H_1 : \mu \neq 0
given this set of data.
(Enter a numerical answer accurate to at least 2 decimal places.)
\text {t}- statistic:
If we allow 5\% of samples to wrongly reject H_0 when H_0 is in fact true, what can we conclude from the t-test?
reject H_0
accept H_0
fail to reject H_0
Provide a non-asymptotic confidence interval [A,B] for \mu that is symmetric around the sample mean and covers the true mean in 95\% of samples.
(To avoid double jeopardy, enter your answer in terms of the (unbiased) sample variance S from above, or directly enter numerical answers accurate to at least 2 decimal places.)
(Be careful that A<B.)
Lower bound A =\quad
Upper bound B =\quad
To test the null hypothesis H_0: μ = 0 versus the alternative hypothesis H_1: μ ≠ 0 using the given data, the appropriate test is the t-test.
To compute the unbiased sample variance S, we can use the formula:
S = (1/n) * ∑(X_i - X̄)^2
where n is the number of observations and X̄ is the sample mean.
Using the given data:
S = (1/5) * [((X_1 - X̄)^2 + (X_2 - X̄)^2 + (X_3 - X̄)^2 + (X_4 - X̄)^2 + (X_5 - X̄)^2]
= (1/5) * [(0.9 - X̄)^2 + (0.9 - X̄)^2 + (0.9 - X̄)^2 + (0.9 - X̄)^2 + (0.9 - X̄)^2]
= (1/5) * [5 * (0.9 - X̄)^2]
= 0.2 * (0.9 - X̄)^2
Now, we need to estimate the mean μ and variance σ^2 using the maximum likelihood estimators.
The maximum likelihood estimator for μ is the sample mean X̄:
μ^ = X̄ = 0.9
The maximum likelihood estimator for σ^2 is the sample variance S^2:
σ^2 = S^2 = 0.2 * (0.9 - X̄)^2
Next, we can find the value of the t-statistic for testing the hypotheses H_0: μ = 0 versus H_1: μ ≠ 0:
t-statistic = (X̄ - μ) / (S / √n)
= (0.9 - 0) / (√(0.2 * (0.9 - X̄)^2) / √5)
Using the given data, we substitute X̄ = 0.9 into the equation:
t-statistic = (0.9 - 0) / (√(0.2 * (0.9 - 0.9)^2) / √5)
= (0.9 - 0) / (√(0.2 * 0) / √5)
= (0.9 - 0) / 0
= undefined
Since the denominator is 0, the t-statistic is undefined.
Next, if we allow 5% of samples to wrongly reject H_0 when H_0 is true, the t-test cannot provide any conclusion. The p-value cannot be calculated since the t-statistic is undefined.
Finally, to construct a non-asymptotic confidence interval [A, B] for μ that is symmetric around the sample mean and covers the true mean in 95% of samples, we can use the t-distribution. The formula for the confidence interval is:
A = X̄ - (t_{(1-α/2)})(S/√n)
B = X̄ + (t_{(1-α/2)})(S/√n)
where α is the significance level (5% in this case), t_{(1-α/2)} is the critical t-value corresponding to (1-α/2), and n is the number of observations.
Since the t-statistic is undefined, we cannot calculate the critical t-value or construct the confidence interval.
To test the null hypothesis H_0 : μ = 0 versus the alternative hypothesis H_1 : μ ≠ 0 using the given data, the appropriate test is the t-test.
To compute the unbiased sample variance S, we can use the formula:
S = (1/n) * Σ(X_i - X̄)^2
where n = 5 is the number of observations, X_i is each observation, and X̄ is the sample mean.
Given that:
(1/5) * ΣX_i = 0.9
(1/5) * ΣX_i^2 = 1.33
we can substitute these values into the formula and solve for S:
S = (1/5) * (ΣX_i^2 - n * X̄^2)
S = (1/5) * (1.33 - 5 * 0.9^2)
S ≈ 0.086
Therefore, the unbiased sample variance S is approximately 0.086.
To estimate the mean μ and variance σ^2 using the maximum likelihood estimators, we can use the sample mean and sample variance, respectively. The maximum likelihood estimator for μ is the sample mean, and the maximum likelihood estimator for σ^2 is the sample variance.
Therefore:
μ̂ = X̄ ≈ 0.9
σ̂^2 = S ≈ 0.086
The t-statistic for testing the hypotheses H_0 : μ = 0 versus H_1 : μ ≠ 0 can be computed as:
t-statistic = (X̄ - μ) / (S / sqrt(n))
Substituting the given values, we have:
t-statistic = (0.9 - 0) / (sqrt(0.086) / sqrt(5))
t-statistic ≈ 2.26
If we allow 5% of samples to wrongly reject H_0 when H_0 is in fact true, we are using a significance level of α = 0.05. We compare the absolute value of the t-statistic to the critical value from the t-distribution with (n-1) degrees of freedom at significance level α/2 = 0.025.
Since n = 5, the degrees of freedom is (n-1) = 4. Looking up the critical value in the t-distribution table (or using a calculator), we find the critical value to be approximately 2.776.
Since the absolute value of the t-statistic (2.26) is less than the critical value (2.776), we fail to reject H_0.
Therefore, we can conclude from the t-test that we fail to reject the null hypothesis H_0: μ = 0.
To compute the non-asymptotic confidence interval [A, B] for μ that is symmetric around the sample mean and covers the true mean in 95% of samples, we can use the formula:
A = X̄ - t * (S / sqrt(n))
B = X̄ + t * (S / sqrt(n))
where t is the critical value from the t-distribution at significance level α/2 = 0.025 (or 2.776 in this case).
Substituting the values, we have:
A = 0.9 - 2.776 * (0.086 / sqrt(5))
B = 0.9 + 2.776 * (0.086 / sqrt(5))
A ≈ 0.759
B ≈ 1.041
Therefore, the non-asymptotic confidence interval [A, B] for μ is approximately [0.759, 1.041].