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Tasya [4]
3 years ago
5

Suppose that we want to estimate the mean reading speed of second graders. The random sample of students' reading speeds we choo

se has a mean of 31.6 words per minute and a standard deviation of 2.4 words per minute. For each of the following sampling scenarios, determine which test statistic is appropriate to use when making inference statements about the population mean. (In the table, Z refers to a variable having a standard normal distribution, and t refers to a variable having a t distribution.) Sampling Scenario could use either Zort unclear (1) The sample has size 100, and it is from a non-normally distributed population with a known standard deviation of 2.6. (2) The sample has size 75, and it is from a non-normally distributed population. (3) The sample has size 12, and it is from a population with a distribution about which we know very little. (4) The sample has size 14, and it is from a normally distributed population with a known standard deviation of 2.6. (5) The sample has size 16, and it is from a normally distributed population with unknown standard deviation.
Mathematics
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: reading speed of a second-grader. (words per minute)

From a random sample of second-grader the mean and standard deviation were:

X[bar]= 31.6 words per minute

S= 2.4 words per minute

To study the population mean (μ) you have the following sampling scenarios and need to choose the correct statistic.

<u>Sampling scenarios:</u>

(1) The sample has size 100, and it is from a non-normally distributed population with a known standard deviation of 2.6.

For this scenario you have to remember the Central Limit Theorem. The variable of study has a non-normal distribution but the sample is large enough (As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation) you can approximate the distribution of the sample mean to normal: X[bar]≈N(μ;σ²/n); Then it is valid to use an approximation to the standard normal distribution to make inference statements about the population mean.

Statistic: Z

(2) The sample has size 75, and it is from a non-normally distributed population.

Same explanation as the statement before, except that the population variance is unknown. Since the distribution is already an approximation you can use its estimation (sample variance) for the statistic. (It would be less accurate than using the population variance but still valid)

Statistic: Z

(3) The sample has size 12, and it is from a population with a distribution about which we know very little.

To use a student's t the population needs to have a normal distribution, which is not the case, same goes for the standard normal.

To apply the central Limit Theorem you need a sample size equal or greater to 30, this is not the case.

For this sampling example neither distribution is applicable.

(4) The sample has size 14, and it is from a normally distributed population with a known standard deviation of 2.6.

The population has a normal distribution and the population standard deviation is known. The sample size is rather small but it isn't an impediment to apply the standard normal distribution. Also, even tough the population standard deviation is known, a student t is also applicable. In this case both statistics are a viable option.

Statistic: Z or t

(5) The sample has size 16, and it is from a normally distributed population with unknown standard deviation.

The population has a normal distribution, with unknown population standard deviation and the sample size is rather small. You can use a Student t to infer over the population mean.

Statistic: t

I hope this helps!

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A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in
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Answer:

h'(t) = -12.6 t +53

Now we can set up the derivate equal to 0 and we have:

-12.6 t +53 = 0

And solving for t we got:

t = \frac{53}{12.6}= 4.206

For the second derivate respect the time we got:

h''(t) = -12.6

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

h(t) = -6.3t^2 +53 t+24

In order to maximize this function we need to take the first derivate respect the time and we have:

h'(t) = -12.6 t +53

Now we can set up the derivate equal to 0 and we have:

-12.6 t +53 = 0

And solving for t we got:

t = \frac{53}{12.6}= 4.206

For the second derivate respect the time we got:

h''(t) = -12.6

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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Answer:

Step-by-step explanation:

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a(x)=x^3-3\\b(x)=2x+1\\c(x)=\frac{3}{x+2} \\d(x)=4-x\\a.\\a(4)=4^3-3=64-3=61\\b.\\\\c(x-5)=\frac{3}{x-5+2} =\frac{3}{x-3} \\c.\\d(-1)=4-(-1)=4+1=5\\b(d(-1))=b(5)=2(5)+1=11\\\\d.\\(coa)(2)=c(a(2))=c(2^3-3)=c(8-3)=c(5)=\frac{3}{5+2} =\frac{3}{7}

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