Question

The TreeDropp’d Fruit Company wants to sell its apples overseas in attractive pink and yellow gift boxes. To design the boxes, the company needs to estimate the average diameter of their apples. A random sample of 50 apples has a mean of 4.1 inches. You assume a population standard deviatino of .4 inches.

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

B. What is the critical z-value for a 95% interval?

C. What is the margin of error for a 95% interval in this scenario? Show your work.

Answer 1

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

`E(\bar X)= (\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)=(1)/(n) n\mu =\mu`

So then yes the
`\bar X[/tex[ is a unbiased estimator for the true mean. </p><p>b) [tex]z_(\alpha/2)=1.96`

c)
`ME= 1.96 (0.4)/(√(50))=0.1109`

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, (\sigma)/(√(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)= (\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)=(1)/(n) n\mu =\mu`

So then yes the
,
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`

Using the normal standard table, excel or a calculator we see that:

`z_(\alpha/2)=1.96`

C. What is the margin of error for a 95% interval in this scenario? Show your work.

The margin of error is given by:

`ME= z_(\alpha/2)(\sigma)/(√(n))`

If we replace the values that we have we got:

`ME= 1.96 (0.4)/(√(50))=0.1109`

The confidence interval on this case is given by:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since we have all the values we can replace:

`4.1 - 1.96(0.4)/(√(50))=3.989`

`4.1 + 1.96(0.4)/(√(50))=4.211`

So on this case the 95% confidence interval would be given by (3.989;4.211)