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Komok [63]
2 years ago
15

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

eneral population. To determine the aptitude of her district's children, a study was conducted. The results of her district's test scores were: 105, 109, 115, 112, 124, 115, 103, 110, 125, 99. If the mean of the general population of school children is 106, what could be said about her claim? Use alpha = .05
Mathematics
1 answer:
anygoal [31]2 years ago
4 0

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1117}{10} = 111.7

Sum of squares of differences = 642.1

S.D = \sqrt{\frac{642.1}{49}} = 8.44

We are given the following in the question:  

Population mean, μ = 106

Sample mean, \bar{x} = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

H_{0}: \mu = 106\\H_A: \mu > 106

We use one-tailed(right) t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833

Since,                  

t_{stat} > t_{critical}

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

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Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

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Solve these arithmetic: 2.75+0.003+0.158 =
olganol [36]
First, let's add 2.75 + 0.158.  First, we add the thousandths places.  Now, we have 2.758 + 0.15.  Next, let's add the hundredths place.  This is equal to 2.808+0.1.  Finally, we have the tenths place.  Now, our number is 2.908.

To finish this problem, we add 0.003.  2.908 + 0.003 = 2.911, so 2.911 is the answer to this problem.
3 0
3 years ago
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Answer:

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We multiply A) by -2

A) -2x -2y  = -16 Then we add this to

B) 3x + 2y = 21

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Second answer is correct

Step-by-step explanation:

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What is a real world problem that could be modeled by dividing two rational numbers
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say, you have half a pie, and 2 sisters, if you divide the half-pie in 3 equal pieces, how much is everyone getting of the whole pie?  well, is the quotient of 1/2 and 3.


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Answer:

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Step-by-step explanation:

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Cost of 1 meter is Rs 17 3/4

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