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1 year ago

What is the standard error of the sampling distribution of sample proportion?.

Mathematics

1 answer:

storchak [24]1 year ago

5 0

Answer:

√ (p(1-p) / n)

Step-by-step explanation:

Standard Error(SE) of the Sample Proportion: √ (p(1-p) / n). Note: as the sample size increases, the standard error decreases.

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Which fractions equal the whole number 4?

Scorpion4ik [409]

Answer:

A, B

Step-by-step explanation:

A. 12/3=4

B. 4/1=4

C. 6/4=1 1/2

D. 4/6=2/3

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2 years ago

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Which of the following options best describes the function graphed below? A graph shows a curve that starts at the origin and go

Tamiku [17]

It's an increasing nonlinear. Exponential functions are graphed in basically any shape besides a line. This is an exponential function, therefore it is classified as nonlinear. By looking at the slope, from left to right, we can see that it is increasing, thus providing an answer that this is an increasing nonlinear function.

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3 years ago

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Chris earns a fixed amount of money every month. At the end of 2 years, chris had earned $86,688. He earned $ each year. He earn

dmitriy555 [2]

Answer:

Chris earned $ 43,344 each year

Monthly Chris earns $ 3,612

Step-by-step explanation:

If you earn a fixed amount of money each month, for 2 years and one year you have 12 months, then 2 years equals 24 months.

To know how much you earn each month we divide the total amount earned in the two years between 24 months

I = $ 86,688 / 24

I = $ 3,612

Monthly Chris earns $ 3,612

To calculate what he earned each year we divided $ 86,688 / 2

I = $ 86,688 / 2

I = $ 43,344

Chris earned $ 43,344 each year

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3 years ago

38=x+10+9-3<br>I don't get this question

Lilit [14]

Answer:

so is that all it says, if its not thats not a question because its already answered

Step-by-step explanation:

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3 years ago

insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of

ss7ja [257]

Answer: $(13.22\%,\ 19.34\%)$

Step-by-step explanation:

Given : Sample space : n= 559

Sample proportion : $\hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628$

$=16.28\%$

Significance level : $\alpha= 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Confidence level for population proportion:-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)$

Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= $(13.22\%,\ 19.34\%)$

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3 years ago