What is the answer of this question?

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

NeX [460]

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

6 0

3 years ago

A ratio compares two values using which operation?

anastassius [24]

The correct answer is division

4 0

3 years ago

Read 2 more answers

A shepherd is paid rs.2400 for gazing 18 cows for 5 weeks.How much will be paid for gazing 20 cows for 8 weeks?

lys-0071 [83]

Answer:

Rs. 4266.67(approx) will be paid for gazing 20 cows for 8 weeks.

Explanation:

A shepherd is paid rs. 2400 for gazing 18 cows for 5 weeks .

⇒ Number of money paid for gazing 1 cow for 1 week is, $\frac{2400}{18\times 5} =\frac{400}{15}= \frac{80}{3}$

then,

for gazing 20 cows for 8 weeks is= Rs. $\frac{80}{3}\times (20\times 8)$

Therefore, the amount paid for gazing 20 cows for 8 weeks is, $\frac{80 \times 160}{3} = \frac{12800}{3} = Rs. 4266.66667$ or Rs. 4266.67(approx)

7 0

3 years ago

Triangle DEF (not shown) is similar to ABC shown, with angle B congruent to angle E and angle C congruent to angle F. The length

Bezzdna [24]

Answer:

Area of ΔDEF is $45\ cm^2$ .

Step-by-step explanation:

Given;

ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Length of AB = $2\ cm$ and

Length of DE = $6\ cm$

Area of ΔABC = $5\ cm^2$

Solution,

Since, ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Therefore,

$\frac{Area\ of\ triangle\ 1}{Area\ of\ triangle\ 2} =\frac{AB^2}{DE^2}$

Where triangle 1 and triangle 2 is ΔABC and ΔDEF respectively.

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

$\frac{5}{Area\ of\ triangle\ 2} =\frac{2^2}{6^2}\\ \frac{5}{Area\ of\ triangle\ 2}=\frac{4}{36}\\ Area\ of\ triangle\ 2=\frac{5\times36}{4} =5\times9=45\ cm^2$