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.
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
1 answer