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pshichka [43]
3 years ago
14

You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 nigh

t students, and the sample mean GPA is 2.35 with a standard deviation of 0.46. You sample 25 day students, and the sample mean GPA is 2.58 with a standard deviation of 0.47. Test the claim using a 5% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Hypotheses:
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.

The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 2.35

x2 = 2.58

s1 = 0.46

s2 = 0.47

n1 = 30

n2 = 25

t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)

t = - 1.82

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862

df = 51

We would determine the probability value from the t test calculator. It becomes

p value = 0.075

Alpha = 5% = 0.05

Since alpha, 0.05 < than the p value, 0.075, then we would fail to reject the null hypothesis.

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Dave and Busters charges 2.50 per game in the arcade plus a one-time entrance fee of $10.
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Answer:

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Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
Alvin is 11 years younger than Elga. The sum of their ages is 43. What is elgas age
Neporo4naja [7]

Answer:

16

Step-by-step explanation:

Lets assume x is Elgas' age

And y is Alvins age

we know that alvin is 11 years younger than Elga which means,

x=y+11

and the sum of their age is 43

x+y=43

solve it and you get 16

5 0
3 years ago
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