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4 years ago

7

for a group experiment, your science class measured the fine-particulate concentration in the air at random places around campus

, and estimated a sample average of 32 micrograms per cubic meter, if 225 readings were take, and the standard deviation of the sample measurement was 3 you are 99.7% confidence that actual concentration of fine particulates at the school is?

2 answers:

allsm [11]4 years ago

7 0

Answer:

$31.4,32.6$

Step-by-step explanation:

In the group experiment, 225 readings were taken, the estimated sample average or mean was 32 micro grams per cubic meter and the standard deviation of the sample measurement was 3.

So here,

n = sample size = 255,

μ = mean = 32,

σ = standard deviation = 3.

We have to find the confidence interval of 99.7% around the mean. For confidence interval of 99.7% z score is 3.

We know that confidence interval is,

$=\mu \pm z\dfrac{\sigma}{\sqrt{n}}$

Putting the values,

$=32 \pm 3\dfrac{3}{\sqrt{225}}$

$=32 \pm 3\dfrac{3}{15}$

$=32 \pm \dfrac{9}{15}$

$=32 \pm 0.6$

$=31.4,32.6$

Rina8888 [55]4 years ago

3 0

Its 31.4 - 32.6 this for apex

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notka56 [123]

Answer:

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Step-by-step explanation:

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3 years ago

At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen

guajiro [1.7K]

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

p2 -> 1997

10 out of 100, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Distribution of p1 – p2:

$p = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Confidence interval:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower bound of the interval is:

$p - zs = 0.1 - 1.645*0.05 = 0.01775
$

The upper bound of the interval is:

$p + zs = 0.1 + 1.645*0.05 = 0.18225
$

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

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3 years ago

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Juli2301 [7.4K]

The answer would be B
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3 years ago

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gtnhenbr [62]

Hey there! :)

To find an equation of a line that passes through (5, 1) and has a slope of 2, we'll need to plug our known variables into the slope-intercept equation.

Slope-intercept equation : y = mx + b ; where m=slope, b=y-intercept

Since we're already given the slope, all we really need to do is find the y-intercept.

We can do this by plugging our known values into the slope-intercept equation.

y = mx + b

Since we're trying to find "b," we need to plug in "y, m, x" into our formula.

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Simplify.

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Subtract 10 from both sides.

1 - 10 = b

Simplify.

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So, our y-intercept is 9!

Now, we can very simply plug our known values into slope-intercept form.

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~Hope I helped!~

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3 years ago