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

For this problem, we have that:

$n = 40361, \pi = \frac{8337}{40361} = 0.2066$

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 $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2066 - 1.645\sqrt{\frac{0.2066*0.7934}{40361}} = 0.2033$

The lower bound for a 90% confidence interval is 0.2033.

6 0

3 years ago

tom has £55,000 in 2016 he plans on putting £8,200 in a bank each year how much will be in the bank by 2031

mr_godi [17]

Answer:

123000

Step-by-step explanation: 8,200 x 15 =123000

7 0

3 years ago