Levelt Question 7 1) What is the total amount of sap the trees produced that day?

Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at colleg

zhenek [66]

Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = average rent paid by 314 BYU students = $346

s = sample standard deviation = $86

n = sample of students = 314

$\mu$ = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, $\mu$ is ;

P(-2.082 < $t_3_1_3$ < 2.082) = 0.96

P(-2.082 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.082) = 0.96

P( $-2.082 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $346 -2.082 \times {\frac{86}{\sqrt{314} } }$ , $346 +2.082 \times {\frac{86}{\sqrt{314} } }$ ]

= [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

5 0

3 years ago

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

What is the length of the hypotenuse of the triangle?

atroni [7]

Step-by-step explanation:

Hey there!

Here;

ABC is a Right angled triangle.

Taking reference angle as angleA.

Now;

Hypotenuse (h) = AB

Perpendicular (BC)= 24 ft.

Base(AC)= 10 ft.

Now, Using Pythagoras relation;

$h = \sqrt{ {p}^{2} + {b}^{2} }$

Put all values.

$h = \sqrt{ {24}^{2} + {10}^{2} }$

Simplify it.

$h = \sqrt{576 + 100}$

$h = \sqrt{676}$

Therefore, h= 26ft.

ANS: OPTION A.

Hope it helps....

3 0

3 years ago

Read 2 more answers

The price of a miniature toy car in 1968 was 0.295% of its current price. Today the car is a collector's item and is priced at $

valentinak56 [21]

$200 x 0.00295
= $0.59

The price of the toy car in 1968 was $0.59

8 0

3 years ago

Read 2 more answers

Least common denominator in 1/30 + 1/24

aev [14]

Answer:

The LCD in 1/30 + 1/24 is 90.

8 0

2 years ago

Levelt Question 7 1) What is the total amount of sap the trees produced that day?​

Levelt Question 7 1) What is the total amount of sap the trees produced that day?