Refer to the following data: Period Demand 1 58 2 59 3 60 4 61 What is the last-value forecast for the next period?

A person recently read that 84% of cat owners are women. How large a sample should the researcher take if she wishes to be 90% c

Lana71 [14]

Answer:

405

Step-by-step explanation:

To find sample size, use the following equation, where n = sample size, za/2 = the critical value, p = probability of success, q = probability of failure, and E = margin of error.

$n=\frac{(z_{\frac{\alpha }{2} })^{2} *p*q }{E^2}$

The values that are given are p = 0.84 and E = 0.03.

You can solve for the critical value which is equal to the z-score of (1 - confidence level)/2. Use the calculator function of invNorm to find the z-score. The value will given with a negative sign, but you can ignore that.

(1 - 0.9) = 0.1/2 = 0.05

invNorm(0.05, 0, 1) = 1.645

You can also solve for q which is 1 - p. For this problem q = 1 - 0.84 = 0.16

Plug the values into the equation and solve for n.

$n =\frac{(1.645)^2*0.84*0.16}{(0.03)^2}\\n= 404.0997333$

Round up to the next number, giving you 405.

6 0

2 years ago

The blue segment below is a radius of O. What is the length of the diameter of the circle?

Ivahew [28]

Answer:

B, 14.6 units.

Step-by-step explanation:

The answer is be because:

the diameter = radius times 2

radius= 7.3

diameter=7.3*2

=14.6

Therefore the answer is, B

7 0

2 years ago

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A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.22, n = 425$

So

$\mu = E(X) = np = 425*0.22 = 93.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

$Z = -1.35$ has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0

2 years ago

In a 3M Privacy Filters poll, 806 adults were asked to identify their favorite seat when they fly, and 492 of them chose a windo

Snowcat [4.5K]

Answer:

Step-by-step explanation:

If majority of adults prefer a window seat when they fly, it means that more than half or more than 0.5 of the population prefer a window seat when they fly.

We would set up the hypothesis test.

For the null hypothesis,

p = 0.45

For the alternative hypothesis,

p > 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 492

n = number of samples = 806

P = 492/806 = 0.61

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.61 - 0.5)/√(0.5 × 0.5)/806 = 6.24

From the normal distribution table, the p value < 0.00001

Since alpha, 0.01 < than the p value, then we would reject the null hypothesis. Therefore, at a 1% significance level, we can conclude that the majority of adults prefer a window seat when they fly.

8 0

2 years ago

Randy collected 30 signatures from fifth grade students at his school. He noticed that 3/5 of the signatures were done in pencil

inna [77]

Answer:

18 signatures

Step-by-step explanation:

Take the total number of signatures and multiply by the fraction that were done in pencil to determine the number done in pencil

30 * 3/5

90/5

18

3 0

2 years ago

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