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blsea [12.9K]
3 years ago
6

What is the answer to this problem?

Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0
The answer to this problem is -25
Lerok [7]3 years ago
5 0
-25 i used google :D
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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors
GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that n = 1600, \pi = \frac{8}{1600} = 0.005

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

A sample of n is required, and n is found for M = 0.009. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}

0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}

\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}

(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2

n = 407.3

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.009\sqrt{n} = 2.575*0.5

\sqrt{n} = \frac{2.575*0.5}{0.009}

(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2

n = 20464.9

Rounding up:

A sample of 20465 is required.

8 0
2 years ago
Find the amplitude of y = -4 sin x.
lyudmila [28]
Answer:
4
Explanation:
The amplitude of a sine function is the absolute value of the coefficient of sin(in other words, the absolute value of the number multiplying the sin)
In this case, that is |-4|=4
6 0
2 years ago
Chile I don’t know this
jolli1 [7]
The answer is 39 degrees
6 0
3 years ago
A bag contains 8 red marbles, 4 white marbles, and 6 blue marbles. find P(red or blue)
Charra [1.4K]

Answer:

7/9 or ~77.7778% or 0.777778

Step-by-step explanation:

First you need to find the total amount or marbles.

  • 8+4+6=18

Next, find the sum of <em>just</em><em> </em>the<em> </em>blue<em> </em>and red marbles.

  • 8+6=14

Now, find the probability by dividing or simplifying the number of blue and red by the total.

  • 14/18= <u>7/9</u> OR <u>~</u><u>0</u><u>.</u><u>7</u><u>7</u><u>7</u><u>7</u><u>7</u><u>8</u>
3 0
3 years ago
Your mom opens an account to save money for a family vacation. The account earns an annual interest rate of 4%. She earns $37 in
Veronika [31]
The formula would be I=(37)(.04)(6) which would give you $8.88
6 0
3 years ago
Read 2 more answers
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