Mathematics, 21.06.2019 15:00 LtPeridot

The head librarian at the library of congress has asked her assistant for an interval estimate of the mean number of books checked out each day. the assistant provides the following interval estimate: from 740 to 920 books per day. what is an efficient, unbiased point estimate of the number of books checked out each day at the library of congress?

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Answer from: vavzincak

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

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".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =150$ represent the population standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

$\bar X = \frac{740+920}{2} = 830$

and the margin of error is given by :

$ME = \frac{920-740}{2}= 90$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

Answer from: PinkDivaGirl02

answer:

step-by-step explanation:

Answer from: grilloj72ovtlxv

At 95% confidence level, she used 11 people to estimate the confidence interval

Step-by-step explanation:

The bounds of the confidence interval are: 740 to 920

Mean is calculated as the average of the lower and upper bounds of the confidence interval. So, for the given interval mean would be:

$u=\frac{740+920}{2}=830$

Margin of error is calculated as half of the difference between the upper and lower bounds of the confidence interval. So, for given interval, Margin of Error would be:

$E=\frac{920-740}{2}=90$

Another formula to calculate margin of error is:

$E=z\frac{\sigma}{\sqrt{n}}$

Standard deviation is given to be 150. Value of z depends on the confidence level. Confidence Level is not mentioned in the question, but for the given scenario 95% level would be sufficient enough.

z value for this confidence level = 1.96

Using the values in above formula, we get:

$90=1.96 \times \frac{150}{\sqrt{n} }\\\\ n = (\frac{1.96 \times 150}{90})^{2}\\\\ n=11$

So, at 95% confidence level her assistant used a sample of 11 people to determine the interval estimate

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