Someone help and please make sure the answer is right :)

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 $\alpha/2=0.025$

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:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

If we replace the values that we have we got:

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

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since we have all the values we can replace:

$4.1 - 1.96\frac{0.4}{\sqrt{50}}=3.989$

$4.1 + 1.96\frac{0.4}{\sqrt{50}}=4.211$

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

4 0

3 years ago

An entertainment reporter examines the average ticket prices of Broadway shows, comparing them to the number of total performanc

Marizza181 [45]

Answer: D

Step-by-step explanation: I believe it's D because there's a STRONG, POSITIVE RELATIONSHIP so they're both increasing, but there's no correlation

3 0

3 years ago

Gold in the form of solid cylinders will be molded into solid blocks. Each cylinder has a diameter of 6.19 cm and a height of 6

Taya2010 [7]

First, solve for the volume (V) of each cylinder by the equation,
V = πd²h / 4
where d and h are diameter and height, respectively. Substituting the given values,
V = π(6.19 cm)² x (6 cm) / 4 = 180.56 cm³
To determine the number of cylinders needed, divide the volume of the solid block by the volume of the cylinder,
n = 360 cm³ / 180.56 cm³ = 1.99
Thus, approximately 2 cylinders are needed to mold one block of gold.

5 0

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11). Liam, Lidea, Lisa and Larry bought a

denis-greek [22]

Answer:Sorry about the sick day.The answer is simply Larry did the greatest.

Step-by-step explanation:First, subtract 54-14=41. Then, you multiply 41 times .25 since 25%=.25.This will give you 10.25.Finally, you subtract 41.00-10.25 to find out the remaining payment from Larry which is 37.50 dollars.Therefore, Larry paid the most.

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Helpppppppppppppppppppppppppp

Ymorist [56]

She has to add 2 more quarters for 108 more days.

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