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Lelu [443]
3 years ago
9

Evaluate how organizations can use one-sample hypothesis testing to determine if there are performance issues in the organizatio

n. Support your response with a specific example.
Mathematics
1 answer:
Margarita [4]3 years ago
3 0

Answer:

Organizations can use one-sample hypothesis test to determine if there are performance issues in many ways.

It can be applied to the performance of a sector, a machine, a product, an advertising campaing, etc.

For example, we can take the example of a machine. It may be claimed that a specific machine performs significantly worse than the average.

This average would be the population mean: the average performance of the machines of the same type or process.

Then, a sample of the performance of the machine in study is taken and the hypothesis test can be performed to test the claim that this machine performs significantly worse.

Step-by-step explanation:

For example, we have an historic performance for this type of machine of 100 units a day. The machine A in study is sampled 14 days and have a performance of 92 units a day, with a sample standard deviation of 12 units/day. We have to test the claim that the machine A makes less units per day than the average.

Then, the null and alternative hypothesis are:

H_0: \mu=100\\\\H_a:\mu< 100

The significance level is 0.05.

The sample has a size n=14.

The sample mean is M=92.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{14}}=3.2071

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{92-100}{3.2071}=\dfrac{-8}{3.2071}=-2.4944

The degrees of freedom for this sample size are:

df=n-1=14-1=13

This test is a left-tailed test, with 13 degrees of freedom and t=-2.4944, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.0134) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that machine A produces significantly less units per day than the average.

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The legs of a isosceles right triangle both measure 10 inches find the length of the hypotenuse
Artist 52 [7]

Answer:

Hypotenuse = 14.1421 inches (rounded off to four decimal values)

Step-by-step explanation:

The legs of an isosceles triangle both measure 10 inches

Since the triangle is a right triangle,

We apply the Pythagorean Theorem to find the height of the hypotenuse

a² + b² = c²

a = 10 inches, b = 10 inches and (c) is the hypotenuse whose value we don't know.

So; 10² + 10² = c²

c² = 100 + 100

c² = 200

c = \sqrt{200} = 14.1421356237 inches

The hypotenuse is 14.1421 inches (answer rounded off to four decimal values)

7 0
3 years ago
Can someone help me??
Allushta [10]

Answer:

67.5 degrees

Step-by-step explanation:

The piece that isn't marked with xs is 90 degrees marked by the square symbol representing a right angle.

Full circle = 360 degrees

360 degrees - 90 degrees = 270 degrees

Now, if we add all the like terms we have 4 x's in total. This is becauswe 3x + another x = 4x.

So, we need to divide 270 by 4. We get 67.5 degrees.

Hope this helps!

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4 0
3 years ago
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Many professional schools require applicants to take a standardized test.
Mars2501 [29]

Answer:

if pete placed in the 73rd percentile, this means that his score was better than 73% of the other test takers, so the fourth option: he scored more than about 73% of the test takers

Step-by-step explanation:

hope this helps :)

5 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
3
Elanso [62]

Answer:

Step-by-step explanation:

6 0
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