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 $1 - \frac{\alpha}{2}$ .

A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.

This means that $n = 65, \pi = 0.76$

99% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.6236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.8964$

0.6236*100 = 62.36%

0.8964*100 = 89.64%

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

6 0

3 years ago

For Thanksgiving Break Nykouth and her family is going to visit relatives in Nebraska. Her dad has found a hotel room at the Hil

taurus [48]

Answer:

109n + 35

Step-by-step explanation:

I'm not sure what you are asking because you have no question in there

8 0

2 years ago

Help me, please Which statement is true about the following pair of rectangles?

miv72 [106K]

Answer:

Step-by-step explanation:

$\frac{27}{33} = \frac{9}{11}$

5 0

3 years ago