Whats the congruence of these two triangles ? none ? sas ? asa ? or sss ?

krok68 [10]

The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

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To solve this question, we need to find the confidence interval for the amount of time it takes the students to get the degree.

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 80 - 1 = 79

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95% confidence interval

Standard level of confidence, we have to find a value of T, which is found looking at the t table, with 79 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9905.

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The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9905\frac{2.2}{\sqrt{80}} = 0.5$

In which s is the standard deviation of the sample and n is the size of the sample.

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The lower end of the interval is the sample mean subtracted by M. So it is 4.8 - 0.3 = 4.3 years.

The upper end of the interval is the sample mean added to M. So it is 4.8 + 0.3 = 5.3 years.

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The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

A similar question is given at brainly.com/question/24278748

6 0

3 years ago

What is -2-5(-2-7i) simplify with full working?

Valentin [98]

Please give thanks and brainliest if correct!

8 0

3 years ago

A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We wan

svlad2 [7]

Answer:

$t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$

Step-by-step explanation:

We have a sample of size n = 35, $\bar{x} = 92$ and $s = 5$ . As we want to construct a 95% confidence interval to estimate the value of the true population mean $\mu$ , we should use the pivotal quantity given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value $t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$ , i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that $P(-2.0322\leq T\leq 2.0322) = 0.95$ because of the simmetry of the t distribution.