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Naddika [18.5K]
3 years ago
12

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time).

The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars. Private Colleges 51.8 42.2 45.0 34.3 44.0 29.6 46.8 36.8 51.5 43.0 Public Colleges 20.3 22.0 28.2 15.6 24.1 28.5 22.8 25.8 18.5 25.6 14.4 21.8.(a)Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for private colleges. (Round the standard deviation to two decimal places.)
sample mean $ thousand sample standard deviation $ thousand
Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for public colleges. (Round the standard deviation to two decimal places.)
sample mean $ thousand sample standard deviation $ thousand
(b)What is the point estimate (in thousand dollars) of the difference between the two population means? (Use Private − Public.)
$ thousandInterpret this value in terms of the annual cost (in dollars) of attending private and public colleges.
We estimate that the mean annual cost to attend private colleges is $ more than the mean annual cost to attend public college
(c)Develop a 95% confidence interval (in thousand dollars) of the difference between the mean annual cost of attending private and public colleges. (Use Private − Public. Round your answers to one decimal place.)$ thousand to $ thousand
Mathematics
1 answer:
NeX [460]3 years ago
6 0

Answer:

(a) PRIVATE COLLEGES

Sample mean is $42.5 thousand

Sample standard deviation is $6.65 thousand

PUBLIC COLLEGES

Sample mean is $22.3 thousand

Sample standard deviation is $4.34 thousand

(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)

(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand

Step-by-step explanation:

(a) PRIVATE COLLEGES

Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand

Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand

PUBLIC COLLEGES

Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand

Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand

(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.

This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges

(c) Confidence Interval = Mean + or - Margin of error (E)

E = t×sd/√n

Mean = $42.5 - $22.3 = $20.2 thousand

sd = $6.65 - $4.34 = $2.31 thousand

n = 10+12 = 22

degree of freedom = 22-2 = 20

t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086

E = 2.086×$2.31/√22 = $1.0 thousand

Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand

Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand

95% confidence interval is $19.2 thousand to $21.2 thousand

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HEEELP
Drupady [299]

Answer: (c)

Step-by-step explanation:

Given

f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}

Here, \sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)

To get f\left(g(x)\right), replace x in f(x) by g(x)\ \text{i.e. by}\ \sqrt{x+5}

\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)

Using (i) and (ii)  it can be concluded that the domain of f\left(g(x)\right) is all real numbers except 0.

Therefore, its domain is given by

x\in [-5,4)\cup (4,\infty)

Option (c) is correct.

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