Construct a 90% confidence interval of the population proportion using the given information.

X= 175

n= 250

Tbe confidence interval marked the range of values for which the true population mean is estimated to be given a certain level of confidence. Hence, the confidence interval is (0.6523, 0.7477)

Since the sample size is large enough, we use the the Z distributuon :

Confidence interval is defined thus :

Margin of Error :

Hence,

Lower confidence boundary = 0.7 - 0.0290 = 0.6523

Upper confidence boundary = 0.7 + 0.0290 = 0.7477

Therefore, the confidence interval is (0.6523, 0.7477)

Learn more : link

The 90% confidence interval of the population proportion is (0.6523, 0.7477).

Step-by-step explanation:

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

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval of the population proportion is (0.6523, 0.7477).

the first one, all those points on the line just make a strait line.