The workers at sandbachian, inc. took a random sample of 800 manhole covers and found that 40 of them were defective. what is the 95% ci for p, the true proportion of defective manhole covers, based on this sample? a) (37.26, 42.74)b) (.035, .065)c) (.047, .053)d) (.015, .085)

b)

Step-by-step explanation:

The confidence interval for proportion (p) is given by :-

, where = Sample proportion

n= sample size.

z* = Critical z-value.

Let p be the true proportion of defective manhole covers, based on this sample.

Given : The workers at Sandbachian, Inc. took a random sample of 800 manhole covers and found that 40 of them were defective.

Then , n= 800

Confidence interval = 95%

We know that the critical value for 95% Confidence interval : z*=1.96

Then, the 95% CI for p, the true proportion of defective manhole covers will be :-

Hence, the required confidence interval : b)

im pretty sure it would be 5 am, hope that , and give brainliest and guys if this is right then press the button and give 5 stars

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!