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xxTIMURxx [149]
3 years ago
7

The mean annual tuition and fees for a sample of 15 private colleges was with a standard deviation of . A dotplot shows that it

is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from 32,500 a) state the null and alternate hypotheses b) calculate the standard error c) calculate the test statistic d) find the p - value .
Mathematics
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

The mean annual tuition and fees for a sample of 15 private colleges was $35,500 with a standard deviation of $6500. A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from $32,500. State the null and alternate hypotheses. A) H0: 4 = 32,500, H:4=35,500 C) H: 4 = 35,500, H7:35,500 B) H: 4 = 32,500, H : 4 # 32,500 D) H0:41 # 32,500, H : 4 = 32,500

Solution

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 32500

For the alternative hypothesis,

Ha: µ ≠ 32500

This is a two tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 15,

Degrees of freedom, df = n - 1 = 15 - 1 = 14

t = (x - µ)/(s/√n)

Where

x = sample mean = 35500

µ = population mean = 32500

s = samples standard deviation = 6500

t = (35500 - 32500)/(6500/√15) = 1.79

We would determine the p value using the t test calculator. It becomes

p = 0.095

Assuming alpha = 0.05

Since alpha, 0.05 < than the p value, 0.095, then we would fail to reject the null hypothesis.

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Answer:

See explanation below

Step-by-step explanation:

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4 0
3 years ago
Solve the equation by completing the square. Round to the nearest hundreth if necessary. x^2-x-7=0
gregori [183]
X^2 - x  = 7
(x - 0.5)^2 - 0.25 = 7
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solution set is {-2.19, 3.19,}
8 0
3 years ago
Help me with math please
pentagon [3]

Answer:

B. 1/6

Step-by-step explanation:

8 0
3 years ago
The radius of the base of a cylinder is increasing at a rate of 7 millimeters per hour. The height of the cylinder is fixed at 1
Ilya [14]

Answer:

The rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

Step-by-step explanation:

Given:

Rate of increase of base of radius of base of cylinder = 7 mm/hr

Height of cylinder = 1.5 mm

Radius at a certain instant = 12 mm

To find rate of change of volume of cylinder at that instant.

Solution:

Let r represent radius of base of cylinder at any instant.

Rate of increase of base of radius of base of cylinder can be given as:

\frac{dr}{dt}=7\ mm/hr

Volume of cylinder is given by:

V=\pi\ r^2h

Finding derivative of the Volume with respect to time.

\frac{dV}{dt}=\pi\ h\ 2r\frac{dr}{dt}

Plugging in the values given:

\frac{dV}{dt}=\pi\ (1.5)\ 2(12)(7)

\frac{dV}{dt}=252\pi

Using \pi=3.14

\frac{dV}{dt}=252(3.14)

\frac{dV}{dt}=791.28\ mm^3/hr (Answer)

Thus rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

6 0
3 years ago
A right cylinder has a diameter of 40 mm and a height of 30 mm, as shown. Semester Test, Part 1 What is the volume of the cylind
sukhopar [10]
Volume = Area * height 
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 = (pi/4)*(40)^2 * 30 
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so, the answer is (a)
8 0
3 years ago
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