Complete the work shown to answer the question.

A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she

Nataly_w [17]

Answer:

The minimum sample size needed is $n = (\frac{1.96\sqrt{\sigma}}{4})^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

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

$4 = 1.96\frac{\sigma}{\sqrt{n}}$

$4\sqrt{n} = 1.96\sqrt{\sigma}$

$\sqrt{n} = \frac{1.96\sqrt{\sigma}}{4}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{\sigma}}{4})^2$

$n = (\frac{1.96\sqrt{\sigma}}{4})^2$

The minimum sample size needed is $n = (\frac{1.96\sqrt{\sigma}}{4})^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.

4 0

3 years ago

What is the slope intercept of 6x-2y=10

Archy [21]

Answer:

y=3x-5

Step-by-step explanation:

slope intercept form: y=mx+b

-2y=-6x+10

2y=6x-10

y=3x-5

3 0

3 years ago

A store is instructed by corporate headquarters to put a markup of 25% on all items. An item costing $24 is displayed by the s

larisa [96]

Answer: The correct selling price is $29.97.

Step-by-step explanation:

Since we have given that

Cost price of an item = $27

Mark up rate = 11%

So, Amount of mark up would be

So, Amount after mark up would be

Hence, the correct selling price is $29.97.

The manager's likely error is that he has put the selling price the mark up amount only i.e $2.9≈$3 instead of adding the mark up amount to the cost price.

5 0

3 years ago

Read 2 more answers

Y=-4x.-3 y= -2x + 1 What is y and x?

Jet001 [13]

You would equal the equations together then solve for a variable.

-4x - 3 = -2x + 1

-4x + 2x = 1 + 3

-2x = 4
—— —-
-2 -2

x = -2

then take that x value and substitute it into any one of the equations.

Equation #1
y = -4(-2) -3
= 8 - 3
= 5

Equation #2
y = -2(-2) + 1
= 4 + 1
= 5

4 0

3 years ago

Read 2 more answers

Confidence Interval Mistakes and Misunderstandingsâ€”Suppose that 500 randomly selected recent graduates of a university were as

kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

$n=500\\\bar x=8.6\\\sigma=2.2$

(a)

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

$8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)$

(b)

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

n = 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0

3 years ago