Find an interval I𝜃 (that depends on 𝜃 ) centered about 𝑋⎯⎯⎯⎯⎯𝑛 such that

Question

Find an interval I𝜃 (that depends on 𝜃 ) centered about 𝑋⎯⎯⎯⎯⎯𝑛 such that

𝐏(I𝜃∋𝜃)=0.9for all 𝑛(i.e, not only for large 𝑛).

(Write barX_n for 𝑋⎯⎯⎯⎯⎯𝑛 . Use the estimate 𝑞0.05≈1.6448 for best results.)

I𝜃=[𝐴𝜃,𝐵𝜃] for

𝐴𝜃 =

𝐵𝜃 =

Answer 1

1. TRUE

2. mean = theta
variance = theta/n

Answer 2

To find the interval I𝜃 centered around 𝑋⎯⎯⎯⎯⎯𝑛 such that 𝐏(I𝜃∋𝜃) = 0.9, we can use the normal distribution and the Z-score.

The Z-score is calculated using the formula:

Z = (𝜃 - 𝑋⎯⎯⎯⎯⎯𝑛) / 𝜎

where 𝑋⎯⎯⎯⎯⎯𝑛 is the sample mean, 𝜃 is the true mean, and 𝜎 is the standard deviation.

Since we want to find the interval I𝜃, we need to find the values 𝐴𝜃 and 𝐵𝜃 that represent the lower and upper bounds of the interval, respectively.

The probability 𝐏(I𝜃∋𝜃) = 0.9 implies that the area under the normal distribution curve between 𝐴𝜃 and 𝐵𝜃 is 0.9.

To find 𝐴𝜃, we can use the formula:

𝐴𝜃 = 𝑋⎯⎯⎯⎯⎯𝑛 - (𝑞0.05 * 𝜎 / √𝑛)

where 𝑞0.05 is the Z-score corresponding to a 95% confidence level. Using the estimate 𝑞0.05 ≈ 1.6448, we can substitute it into the formula:

𝐴𝜃 = 𝑋⎯⎯⎯⎯⎯𝑛 - (1.6448 * 𝜎 / √𝑛)

To find 𝐵𝜃, we can use the formula:

𝐵𝜃 = 𝑋⎯⎯⎯⎯⎯𝑛 + (𝑞0.05 * 𝜎 / √𝑛)

Substituting 𝑞0.05 ≈ 1.6448 into the formula:

𝐵𝜃 = 𝑋⎯⎯⎯⎯⎯𝑛 + (1.6448 * 𝜎 / √𝑛)

Therefore, the interval I𝜃 is given by:

I𝜃 = [𝐴𝜃, 𝐵𝜃] = [𝑋⎯⎯⎯⎯⎯𝑛 - (1.6448 * 𝜎 / √𝑛), 𝑋⎯⎯⎯⎯⎯𝑛 + (1.6448 * 𝜎 / √𝑛)]

Please note that the exact values of 𝐴𝜃 and 𝐵𝜃 will depend on the specific values of 𝑛, 𝑋⎯⎯⎯⎯⎯𝑛, and 𝜎.

Answer 3

To find an interval I𝜃 that satisfies 𝐏(I𝜃∋𝜃)=0.9 for all 𝑛, we can use the concept of confidence intervals.

A confidence interval is an interval estimate that expresses the uncertainty surrounding the value of a population parameter. In this case, we are interested in estimating the value of 𝜃 with 90% confidence.

To construct the confidence interval, we need to consider the mean 𝑋⎯⎯⎯⎯⎯𝑛 and the standard deviation 𝜎 of the population.

The formula for a confidence interval, when the population standard deviation is known, is given by:

𝐶𝐼 = [𝑋⎯⎯⎯⎯⎯𝑛 − 𝑧(𝛼/2) * 𝜎/√𝑛, 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑧(𝛼/2) * 𝜎/√𝑛]

In this formula, 𝑧(𝛼/2) is the z-score associated with the desired level of confidence. Given that 𝑞0.05 ≈ 1.6448, we can use this value.

However, in our case, the population standard deviation 𝜎 is not known. So, we need to estimate it using the sample standard deviation 𝑆𝑛.

The formula for the confidence interval, when the population standard deviation is unknown, is given by:

𝐶𝐼 = [𝑋⎯⎯⎯⎯⎯𝑛 − 𝑡(𝛼/2, 𝑛−1) * 𝑆𝑛/√𝑛, 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑡(𝛼/2, 𝑛−1) * 𝑆𝑛/√𝑛]

Here, 𝑡(𝛼/2, 𝑛−1) represents the t-score associated with the desired level of confidence and the sample size 𝑛.

Since we want this interval to hold true for all 𝑛, we need to ensure that the width of the interval remains constant as 𝑛 changes. Therefore, we will take the largest possible value of 𝑆𝑛, which occurs when 𝑛 is smallest, i.e., 𝑛 = 1.

For 𝑛 = 1, the width of the interval is given by:

Width = 2 * 𝑡(𝛼/2, 1-1) * 𝑆1/√1

Since there is only one observation, 𝑡(𝛼/2, 1-1) can be chosen arbitrarily. Let's select 𝑡(𝛼/2, 1-1) = 1 for simplicity.

Therefore, the width of the interval is:

Width = 2 * 1 * 𝑆1/√1 = 2 * 𝑆1

Now, let's set this width equal to the desired width of the interval, which is equal to 2𝑧(𝛼/2)𝜎/√𝑛:

2 * 𝑆1 = 2 * 𝑧(𝛼/2)𝜎/√𝑛

Simplifying and solving for 𝜎, we get:

𝜎 = 𝑆1 * √𝑛/𝑧(𝛼/2)

Substituting the known values, 𝑆1 = |𝑞0.05| = |1.6448| and 𝑧(𝛼/2) = 𝑞0.05 ≈ 1.6448, we have:

𝜎 = |1.6448| * √𝑛/|1.6448|

Simplifying further, we get:

𝜎 = √𝑛

Now, substitute this value of 𝜎 in the formula for the confidence interval:

𝐶𝐼 = [𝑋⎯⎯⎯⎯⎯𝑛 − 𝑡(𝛼/2, 𝑛−1) * 𝑆𝑛/√𝑛, 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑡(𝛼/2, 𝑛−1) * 𝑆𝑛/√𝑛]

𝐶𝐼 = [𝑋⎯⎯⎯⎯⎯𝑛 − 𝑡(𝛼/2, 𝑛−1) * √𝑛, 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑡(𝛼/2, 𝑛−1) * √𝑛]

Since we want the interval to be centered around 𝑋⎯⎯⎯⎯⎯𝑛, the lower bound 𝐴𝜃 would be 𝑋⎯⎯⎯⎯⎯𝑛 − 𝑡(𝛼/2, 𝑛−1) * √𝑛, and the upper bound 𝐵𝜃 would be 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑡(𝛼/2, 𝑛−1) * √𝑛.

Therefore, the desired interval is:

I𝜃 = [𝐴𝜃, 𝐵𝜃] = [𝑋⎯⎯⎯⎯⎯𝑛 − 𝑡(𝛼/2, 𝑛−1) * √𝑛, 𝑋⎯⎯⎯⎯⎯𝑛 + 𝑡(𝛼/2, 𝑛−1) * √𝑛]

Please note that the specific values of 𝑋⎯⎯⎯⎯⎯𝑛 and 𝑡(𝛼/2, 𝑛−1) depend on the context of your problem. Make sure to substitute the appropriate values to obtain the final interval.