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UNO [17]
3 years ago
7

g The bursar a conducts a test of hypothesis that the mean amount of student debt a WCU undergrad will leave college with is equ

al to $10,000, versus the alternative that it is less than $10,000. The data leads to the rejection the null hypothesis. However, it is learned later that the mean is indeed $10,000. What type of error (if any) occurred
Mathematics
1 answer:
aleksandrvk [35]3 years ago
4 0

Answer: Type I error

Step-by-step explanation:

A type 1 error is also referred to as the false positive and it is when a true null hypothesis is incorrectly rejected by the researcher.

On the other hand, type II error which is also refered to as the false-negative is when a null hypothesis which is false is failed to be rejected by the researcher but rather accepted.

Based on the information given in the question, since the data leads to the rejection of the null hypothesis such was later found out to be true, then a type I error has occured.

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8 - 5 = -3

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3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
Korolek [52]

Answer:

A sample of 499 is needed.

Step-by-step explanation:

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}.

The margin of error is given by:

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

In this question, we have that:

\pi = 0.21

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when M = 0.03. So

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

0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2

n = 498.81

Rounding up

A sample of 499 is needed.

8 0
2 years ago
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