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1 year ago

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The standard deviation of a point estimator is the _____. a. standard error b. sample statistic c. point estimate d. sampling er

r

1 answer:

ivann1987 [24]1 year ago

4 0

The standard deviation of a point estimator is the <u>standard error.</u>

"The point estimate is used to estimate the exact value of any parameter. For example, an internet poll says that 45% of Americans are concerned for the environment."

"Point estimates is a good measure but the problem with this parameter is that the sample of the statistics may vary so that there is a chance of sampling error."

"As the researcher is unable to presume that closeness of the sample statistics parameter with respect to the parameter so researcher refers to calculate the standard error instead of point estimate."

To learn more about standard error here

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larisa [96]

Answer:I got the pic over here

Step-by-step explanation:

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2 years ago

A 12 ounce bottle of shampoo lasts Juan 16 weeks. Assuming he uses the same amount each week, write an equation relating the oun

trasher [3.6K]

He uses one and one third ounces of shampoo per week.

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3 years ago

80 students at a particular school are randomly selected for a survey. 52 cat cereal for breakfast. If

rewona [7]

Answer:

715

Step-by-step explanation:

52/80 × 1100=715 students eat cereal for breakfast

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2 years ago

A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place

Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

$P(t) = P(0)(1-r)^{t}$

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

$P(1) = 0.97P(0)$

Then

$P(t) = P(0)(1-r)^{t}$

$0.97P(0) = P(0)(1-r)^{0}$

$1 - r = 0.97$

So

$P(t) = P(0)(0.97t)^{t}$

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.5P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.5$

$\log{(0.97)^{t}} = \log{0.5}$

$t\log{0.97} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.97}}$

$t = 22.76$

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.85P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.85$

$\log{(0.97)^{t}} = \log{0.85}$

$t\log{0.97} = \log{0.85}$

$t = \frac{\log{0.85}}{\log{0.97}}$

$t = 5.34$

5.34 years for the sample to decay to 85% of its original amount

8 0

3 years ago

Simplify . 3 sqrt 10 + 7 sqrt 15 - 6 sqrt 10 -4 sqrt 15

andre [41]

Take the like terms of the equation.

$3 \sqrt{10} - 6 \sqrt{10}$

$7 \sqrt{15} - 4 \sqrt{15}$

Subtract those like terms together.

$3 \sqrt{10} - 6 \sqrt{10} = -3 \sqrt{10}$

$7 \sqrt{15} - 4 \sqrt{15} = 3 \sqrt{15}$

The simplified form of this equation is $-3 \sqrt{10} + 3 \sqrt{15}$ .

6 0

3 years ago