Thirty million three hundred forty three thousand one hundred and fifty four in number form

WINSTONCH [101]

63000 is already in standard form

7 0

3 years ago

The TreeDropp’d Fruit Company wants to sell its apples overseas in attractive pink and yellow gift boxes. To design the boxes, t

SashulF [63]

Answer:

a)For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

And if we assume that each observation $X_1 , X_2,...,X_n$ follows a normal distribution $X_i \sim N(\mu,\sigma)$ then we have:

$E(\bar X)=\frac{1}{n} n\mu =\mu$

So then yes the $\bar X[/tex[ is a unbiased estimator for the true mean. b) [tex]z_{\alpha/2}=1.96$

c) $ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109$

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=4.1$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.4$ represent the population standard deviation

n=50 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.4$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

A. What is the point estimate in this scenario? Is it an unbiased estimator?

For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

And if we assume that each observation $X_1 , X_2,...,X_n$ follows a normal distribution $X_i \sim N(\mu,\sigma)$ then we have:

$E(\bar X)=\frac{1}{n} n\mu =\mu$

So then yes the $\bar X[/tex[ is a unbiased estimator for the true mean. B. What is the critical z-value for a 95% interval?
The next step would be find the value of [tex]\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$