MATHEMATICS COLLEGE

2. In an industrial training program, students have been averaging about 64 points on a standardized test. The lecture system was replaced by teaching machines with a lab instructor. There was some doubt as to whether the scores would decrease, increase, or stay the same. A sample of n = 60 students using the teaching machines was tested, resulting in a mean of 68 and a standard deviation of 12. Perform a hypothesis test to see if scores would decrease, increase, or stay the same. Use α = 0.05. Be sure to:1. State your hypotheses.
2. Find the value of the Test Statistic.
3. Find the p-value
4. State your decision (Reject or not)
5. State your conclusion.

Answers

Answer 1

Answer:

Answer:

Case I

Null hypothesis: $\mu = 64$

Alternative hypothesis: $\mu \neq 64$

$t=(68-64)/((12)/(√(60)))=2.582$

$df=n-1=60-1=59$

Since is a two sided test the p value would given by:

$p_v =2*P(t_((59))>2.582)=0.012$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is different from 64.

Case II

Null hypothesis: $\mu \leq 64$

Alternative hypothesis: $\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_((59))>2.582)=0.006$

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance

Step-by-step explanation:

Data given and notation

$\bar X=68$ represent the sample mean

$s=12$ represent the sample standard deviation

$n=60$ sample size

$\mu_o =64$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the population mean is different from 64 the system of hypothesis are :

Null hypothesis: $\mu = 64$

Alternative hypothesis: $\mu \neq 64$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(68-64)/((12)/(√(60)))=2.582$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=60-1=59$

Since is a two sided test the p value would given by:

$p_v =2*P(t_((59))>2.582)=0.012$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is different from 64.

Now let's assume that we want to see if the mean is significantly higher than 64

Null hypothesis: $\mu \leq 64$

Alternative hypothesis: $\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_((59))>2.582)=0.006$

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance

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In a survey of 1016 ?adults, a polling agency? asked, "When you? retire, do you think you will have enough money to live comfortably or not. Of the 1016 ?surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.A. There is a 99?% probability that the true proportion of worried adults is between ___ and ___.B. 99?% of the population lies in the interval between ___ and ___.C. There is 99?% confidence that the proportion of worried adults is between ___ and ___.

A stock of food is enough to fees 50 persons for 14 days. How many days will the foos last if 20 persons will be added?

Answers

Answer:

solution

Step-by-step explanation:8 days

Determine the circumference of the base of the tin?

Answers

By answering the given question, we may state that We can apply the formula if we know the radius: C = 2πr where the radius r is.

what is diameter?

In geometry, a circle's diameter is any straight line segment whose endpoint is on the circle and which passes through its centre. Another name for it is the circle's longest chord. The diameter of a sphere can be defined using either idea. The diameter is the length of the line perpendicular to the two points at either end of the circle. If you think of length as the distance between two points, then diameter is length. The diameter of a circle is the separation between its two farthest points.

We must know the base's diameter or radius in order to calculate the circumference of the tin.

If we know the diameter, we can apply the following formula to determine the circle's circumference:

C = πd

where d is the diameter and C is the circumference.

We can apply the formula if we know the radius:

C = 2πr

where the radius r is.

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Professors often attempt to determine if the submissions by the students are genuine or copied off the web sources. The program that performs this task is only 95 % accurate in correctly identifying a genuine submission and 80% accurate in correctly identifying copies. Based on the past statistics, 15% of the student turned in copied work. If a work is identified as a copy by the program, what is the probability that it is indeed a sample of copied work.

Answers

Answer:

0.7385 = 73.85% probability that it is indeed a sample of copied work.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = (P(A \cap B))/(P(A))$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Identified as a copy

Event B: Is a copy

Probability of being identified as a copy:

80% of 15%(copy)

100 - 95 = 5% of 100 - 15 = 85%(not a copy). So

$P(A) = 0.8*0.15 + 0.05*0.85 = 0.1625$

Probability of being identified as a copy and being a copy.

80% of 15%. So

$P(A \cap B) = 0.8*0.15 = 0.12$

What is the probability that it is indeed a sample of copied work?

$P(B|A) = (P(A \cap B))/(P(A)) = (0.12)/(0.1625) = 0.7385$

0.7385 = 73.85% probability that it is indeed a sample of copied work.

During a nine-hour snowstorm, it snows at a rate of 2 inches per hour for the first 3 hours, at a rate of 3 inches per hour for the next 5 hours, and at a rate of 0.75 inch per hour for the final hour.How many inches of snow accumulated from the storm?

Answers

Answer:

use f(x)=y=mx+b

let snow = S, time = t instead of y and x

S(t)=mt+b

The rate of inches per hour represents the slope of the graph, m.

The y-variable would be the amount of snow, S.

The x-variable would be the time, t, in hours.

The function has three pieces:

i) S(t)= 2t (slope = 2)

ii) S(t) = 3t (slope = 3)

iii) S(t) = 0.75t (slope = 0.75)

For the first piece, i), t=3, so the amount of snow is 6 inches.

For the second piece, ii) t=5, so the amount of snow is 15 inches.

For the third piece, iii) t=1, so the amount of snow is 0.75 inch.

In total, it snowed 21.75 inches.

total snow

Final answer:

To find the total accumulation of snow during the nine-hour snowstorm, we calculate the snow accumulation for each hour and then sum them up. The total accumulation of snow from the storm is 21.75 inches.

Explanation:

To find the total accumulation of snow during the nine-hour snowstorm, we need to calculate the amount of snow that fell during each hour and then sum them up. First, we calculate the snow accumulation for each hour:

For the first 3 hours, it snowed at a rate of 2 inches per hour, so the accumulation is 3 * 2 = 6 inches.
For the next 5 hours, it snowed at a rate of 3 inches per hour, so the accumulation is 5 * 3 = 15 inches.
For the final hour, it snowed at a rate of 0.75 inch per hour, so the accumulation is 1 * 0.75 = 0.75 inches.

Finally, we sum up the accumulations for each hour: 6 + 15 + 0.75 = 21.75 inches. Therefore, the total accumulation of snow from the storm is 21.75 inches.

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Evaluate ∫C ysin(z)ds, where C is the circular helix given by the equations x = cos(t), y = sin(t), z = t, 0 ≤ t ≤ 2π. SOLUTION The formula for a line integral in space gives the following. ∫y sin(z)ds = sin2(t) dt = (sin(t))2√ (cos(t))2 + (sin(t))2 + 1dt = 1 2 (1 - cos(2t))dt = √2 2 =

Answers

The line integral is

$\displaystyle\int_Cy\sin z\,\mathrm ds=\int_0^(2\pi)y(t)\sin z(t)\,\sqrt{\left((\mathrm dx)/(\mathrm dt)\right)^2+\left((\mathrm dy)/(\mathrm dt)\right)^2+\left((\mathrm dz)/(\mathrm dt)\right)^2}\,\mathrm dt$

We have

$x=\cos t\implies(\mathrm dx)/(\mathrm dt)=-\sin t$

$y=\sin t\implies(\mathrm dy)/(\mathrm dt)=\cos t$

$z=t\implies(\mathrm dz)/(\mathrm dt)=1$

so the integral reduces to

$\displaystyle\int_0^(2\pi)\sin^2t√((-\sin t)^2+\cos^2t+1^2)\,\mathrm dt=\frac{\sqrt2}2\int_0^(2\pi)(1-\cos2t)\,\mathrm dt=\boxed{\frac\pi{\sqrt2}}$

The line integral ∫C ysin(z) ds over the circular helix C, parametrized by x = cos(t), y = sin(t), z = t for 0 ≤ t ≤ 2π, evaluates to π√2.

To evaluate the line integral ∫C ysin(z) ds over the circular helix C given by x = cos(t), y = sin(t), z = t for 0 ≤ t ≤ 2π, we follow these steps:

1. Parameterize the curve: C is already parameterized as x = cos(t), y = sin(t), z = t.

2. Find the differential ds: ds = √(dx² + dy² + dz²) = √(sin²(t) + cos²(t) + 1)dt = √(1 + 1)dt = √2 dt.

3. Evaluate the integral: ∫C ysin(z) ds = ∫[0, 2π] sin(t) * sin(t) * √2 dt = ∫[0, 2π] sin²(t) * √2 dt.

Now, we'll integrate sin²(t) * √2 with respect to t:

∫ sin²(t) * √2 dt = (1/2) * ∫ (1 - cos(2t)) * √2 dt.

Using the power rule for integration, we get:

(1/2) * [(t - (1/2) * sin(2t)) * √2] | [0, 2π].

Plugging in the limits:

(1/2) * [(2π - (1/2) * sin(4π) - (0 - (1/2) * sin(0))) * √2].

Since sin(4π) = sin(0) = 0:

(1/2) * [(2π - 0 - 0) * √2] = π√2.

So, ∫C ysin(z) ds = π√2.

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A basketball player averages 22.5 points scored per game with a standard deviation of 6.2 points. In one game, the number of points the athlete scored was 1.2 standard deviations below his mean. How many points below average was this value?

Answers

Using the normal distribution, it is found that this value was 7.5 points below the average.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = (X - \mu)/(\sigma)$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 22.5, \sigma = 6.2$ .

In one game, the number of points the athlete scored was 1.2 standard deviations below his mean, hence Z = -1.2 and the score was of X, so:

$Z = (X - \mu)/(\sigma)$

$-1.2 = (X - 22.5)/(6.2)$

X - 22.5 = -1.2 x 6.2

X = 15.

15 - 22.5 = 7.5.

This value was 7.5 points below the average.

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Answer:

7.44 is the answer

Step-by-step explanation:

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