Aresearcher wants to estimate the true proportion of people who would buy items they know are slightly defective from thrift shops because of the lower price. after conducting a survey on a sample of 993 persons who regularly shop at thrift stores, he finds that 305 of the individuals would buy a slightly defective item if it cost less than a dollar. calculate the lower bound of a 90% confidence interval for the true proportion of people who would purchase a defective item.

0.283

Step-by-step explanation:

Formula to find the lower limit of the confidence interval for population proportion is given by :-

, where  = sample proportion.

z* = Critical value

n= Sample size.

Let p be the  true proportion of people who would purchase a defective item.

Given : Sample size = 993

Number of individuals would buy a slightly defective item if it cost less than a dollar = 305

Then, sample proportion of people who would purchase a defective item:

Critical value for 90% confidence interval = z*=1.645  (By z-table)

The lower bound of a 90% confidence interval for the true proportion of people who would purchase a defective item will become

 [rounded to the nearest three decimal places.]

Hence, the lower bound of a 90% confidence interval for the true proportion of people who would purchase a defective item.= 0.283

there is no question content here, but you for the free points! : d