CALCU-Question 1 [15 marks]Let X_1, X_2, , X_n be

Question 1 [15 marks] Let

X_{1}, X_{2}, \ldots, X_{n}

be a random sample of independent and identically distributed random variables with

E\left(X_{i}\right)=\mu

and

\operatorname{Var}\left(X_{i}\right)=\sigma^{2}

. Compute the bias of the estimator

\hat{\sigma}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2},

where

\bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i}

.

Solution by Steps

step 1

To find the bias of the estimator

\hat{\sigma}^{2}

, we need to compute

E(\hat{\sigma}^{2}) - \sigma^{2}

step 2

The expected value of the estimator is

E(\hat{\sigma}^{2}) = E\left(\frac{1}{n} \sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\right)

step 3

We expand the term

(X_{i}-\bar{X})^{2}

to

X_{i}^{2} - 2X_{i}\bar{X} + \bar{X}^{2}

step 4

Taking the expectation, we get

E(\hat{\sigma}^{2}) = \frac{1}{n} \sum_{i=1}^{n}(E(X_{i}^{2}) - 2E(X_{i}\bar{X}) + E(\bar{X}^{2}))

step 5

Since

E(X_{i}) = \mu

and

E(X_{i}^{2}) = \operatorname{Var}(X_{i}) + E(X_{i})^{2} = \sigma^{2} + \mu^{2}

, we substitute these into the equation

step 6

We also know that

\bar{X}

is an unbiased estimator of

\mu

, so

E(\bar{X}) = \mu

and

E(\bar{X}^{2}) = \operatorname{Var}(\bar{X}) + E(\bar{X})^{2}

step 7

The variance of

\bar{X}

is

\operatorname{Var}(\bar{X}) = \frac{\sigma^{2}}{n}

because the

X_{i}

are independent and identically distributed

step 8

Substituting the values from steps 5, 6, and 7 into the equation from step 4, we get

E(\hat{\sigma}^{2}) = \frac{1}{n} \sum_{i=1}^{n}(\sigma^{2} + \mu^{2} - 2\mu^{2} + \frac{\sigma^{2}}{n})

step 9

Simplifying the equation, we find

E(\hat{\sigma}^{2}) = \sigma^{2} - \frac{\sigma^{2}}{n}

step 10

The bias of the estimator is then

E(\hat{\sigma}^{2}) - \sigma^{2} = -\frac{\sigma^{2}}{n}

Answer

The bias of the estimator

\hat{\sigma}^{2}

is

-\frac{\sigma^{2}}{n}

.

Key Concept

Bias of an Estimator

Explanation

The bias of an estimator is the difference between the expected value of the estimator and the true value of the parameter being estimated. In this case, the bias is negative, indicating that the estimator

\hat{\sigma}^{2}

tends to underestimate the true variance

\sigma^{2}

.

What is the formula for computing the bias of an estimator in statistics?

Solution by Steps

step 1

Identify the formula for the bias of an estimator

step 2

The bias of an estimator is defined as the difference between the expected value of the estimator and the parameter being estimated

step 3

For the given estimator

\hat{\sigma}^{2}

, the bias is

E(\hat{\sigma}^{2}) - \sigma^{2}

step 4

Calculate the expected value of

\hat{\sigma}^{2}

step 5

The expected value of

\hat{\sigma}^{2}

is

E(\hat{\sigma}^{2}) = \frac{n-1}{n}\sigma^{2}

step 6

Subtract the true parameter

\sigma^{2}

from the expected value to find the bias

step 7

The bias of

\hat{\sigma}^{2}

is

E(\hat{\sigma}^{2}) - \sigma^{2} = \frac{n-1}{n}\sigma^{2} - \sigma^{2}

step 8

Simplify the expression to find the bias

step 9

The bias of

\hat{\sigma}^{2}

is

-\frac{\sigma^{2}}{n}

Answer

The bias of the estimator

\hat{\sigma}^{2}

is

-\frac{\sigma^{2}}{n}

.

Key Concept

Bias of an Estimator

Explanation

The bias of an estimator is the difference between the expected value of the estimator and the true value of the parameter being estimated. For the sample variance estimator

\hat{\sigma}^{2}

, the bias is

-\frac{\sigma^{2}}{n}

, which accounts for the fact that using the sample mean in the variance formula tends to underestimate the population variance.

Question 3 [20 marks] The measurements of the volume (in

\mathrm{mL}

) for 30 samples of cough syrup bottles are summarized below:

\begin{array}{l} 118.3,118.5,118.9,119.0,119.0,119.2,119.6,119.7,119.7,119.8 \\ 119.9,119.9,120.0,120.0,120.1,120.1,120.1,120.2,120.2,120.4 \\ 120.7,120.7,120.8,121.0,121.1,121.3,121.4,121.7,119.8,120.5 \end{array}

Suppose we expect this particular syrup to have a fill volume of

120 \mathrm{~mL}

, and we assume that the population variance is unknown. i) [10 marks] Test whether these samples were taken from the population of this particular syrup. ii) [10 marks] Calculate the 95\% confidence interval for the population mean.

Generated Graph

Solution by Steps

step 1

Calculate the sample mean (

\bar{x}

) of the given volume measurements

step 2

Calculate the sample standard deviation (s) of the given volume measurements

step 3

Use the t-distribution since the population variance is unknown and the sample size is less than 30

step 4

Calculate the t-score using the formula

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}

, where

\mu_0 = 120

mL is the expected fill volume

step 5

Determine the degrees of freedom (df) which is

n - 1

step 6

Find the p-value corresponding to the calculated t-score and degrees of freedom

step 7

If the p-value is less than the significance level (commonly 0.05), reject the null hypothesis that the sample comes from the population with a mean of 120 mL

step 8

Calculate the 95% confidence interval for the population mean using the formula

\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}

step 9

Find the critical t-value (

t_{\alpha/2}

) for a 95% confidence level and

n - 1

degrees of freedom

step 10

Plug the values into the confidence interval formula to get the interval

Answer

[Insert final answer here]

Key Concept

Hypothesis testing and confidence interval estimation using t-distribution

Explanation

When the population variance is unknown and the sample size is small, we use the t-distribution for hypothesis testing and to calculate confidence intervals for the population mean.

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