Thomas invested $1498.00 at 6% simple interest per annum calculate How long it will be before his investment earns $449.40

Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua

Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

$\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2$

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92$

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5$

Then, the probability of getting a average score of 79 or higher is:

$P(X>79)=P(z>-0.5)=0.69146$

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0

3 years ago

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 45 degrees at midnig

Paraphin [41]

Answer:

D = 9sin(2π(t + a)/24) + 45

Step-by-step explanation:

Let's find the average temperature;

(54 + 36)/2 = 45°

Amplitude = 54 - 45 = 9

From the wave equation, we can write the temperature as;

D = 9sin(2π(t + a)/24) + 45

Where;

D is the temperature

t is the time in hours after midnight

a is a "phase" that is used to set the time at which temperature(D) occurs

5 0

3 years ago

A metronome uses an inverted pendulum rod to control tempo. A weight can be slid up the pendulum rod to decrease tempo, or down

Dominik [7]

Answer:

Part 1

Multiply both sides by 2π

2πf = √(g/L)

Square both sides

4π²f²= g/L

Invert both sides

1/(4π²f²) = L/g

Multiply both side by g

g/(4π²f²) = L

We usually write an equation with the subject (L) n the left

L = g/(4π²f²)

________________________

Part 2

Using the above equation with the given values:

L = g/(4π²f²)

. .= 9.8 / (4π² x 1.6²)

. .= 0.097m (= 9.7cm)

________________________

By the way, where it says

“f=1.6 then there are 2 beats a second, or 192 beats per minute (bpm).”

this should say

“f=1.6 then there are 3.2 beats a second, or 192 beats per minute (bpm)”

Step-by-step explanation:

7 0

3 years ago

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor

castortr0y [4]

Answer:

A) 95% confidence interval for the population mean PEF for children in biomass households = (3.214, 3.386)

95% confidence interval for the population mean PEF for children in LPG households

= (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.

C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Step-by-step explanation:

A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.

Finding the critical value from the z-tables,

Significance level for 95% confidence interval

= (100% - 95%)/2 = 2.5% = 0.025

z (0.025) = 1.960 (from the z-tables)

For the children in the biomass households

Sample mean = 3.30

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.20

N = sample size = 755

σₓ = (1.20/√755) = 0.0436724715 = 0.04367

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 3.30 ± (1.960 × 0.04367)

CI = 3.30 ± 0.085598

95% CI = (3.214402, 3.385598)

95% Confidence interval = (3.214, 3.386)

For the children in the LPG households

Sample mean = 4.25

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.75

N = sample size = 750

σₓ = (1.75/√750) = 0.063900965 = 0.063901

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 4.25 ± (1.960 × 0.063901)

CI = 4.25 ± 0.125246

95% CI = (4.12475404, 4.37524596)

95% Confidence interval = (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.

The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.

Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂

The null hypothesis is

H₀: μ ≥ 0 or μ₁ ≥ μ₂

The alternative hypothesis is

Hₐ: μ < 0 or μ₁ < μ₂

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = 3.30

n₁ = 755

s₁ = 1.20

μ₂ = 4.25

n₂ = 750

s₂ = 1.75

σ = √[(1.20²/755) + (1.75²/750)] = 0.07740

z = (3.30 - 4.25) ÷ 0.07740 = -12.27

checking the tables for the p-value of this z-statistic

Significance level = 0.01

The hypothesis test uses a one-tailed condition because we're testing in only one direction.

p-value (for z = -12.27, at 0.01 significance level, with a one tailed condition) = < 0.000000001

The interpretation of p-values is that

When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.

Significance level = 0.01

p-value = 0.000000001

0.000000001 < 0.01

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.

C) For FEY for biomass households,

Sample mean = 2.3 L/s

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation = 0.5

N = sample size = 755

σₓ = (0.5/√755) = 0.0182

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 2.30 ± (1.960 × 0.0182)

CI = 2.30 ± 0.03567

95% CI = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Hope this Helps!!!

6 0

3 years ago

What is the value of (2/7)3?

igor_vitrenko [27]

Is there options for this question

3 0

3 years ago