Economy, asked by srajanjain896, 1 month ago

A random sample of 500 pineapples was drawn from a large consignment out of which 65 were found to be defective. Estimate the proportion of spoiled pineapple in the whole consignment and also find the standard error of the estimate. Prove that the percentage of spoiled pineapples in the entire consignment is definitely found to be between 8.5 and 17.5. A random sample of 500 pine - apples were taken from a large consignment and 65 were found to be bad . Estimate the proportion of the bad pine - apples in the consignment , as well as the standard error of the estimate . Deduce that the percentage of bad pine - apples in the consignment almost certainly lies between 8.5 and 17.5 .

Answers

Answered by khushikumari15122006
14

Answer:

A random sample of 500 pineapples was taken from large consignment of 65 was found to be bad so that the standard error of the population of bad ones in a sample of this size is 0.015 and deduce that the percentage of bad pineapples in the consignment obtained almost certainly lies between 8.5 and 17.5.

Answered by bishaldasdibru
2

Answer :

The percentage of bad pineapples in the consignment lies between 8.6% to 17.4%.

Explanation :

A random sample of 500 pineapples was drawn from a large consignment out of which 65 were found to be defective. To estimate the proportion of spoiled pineapple in the whole consignment, we can use the sample proportion of defective pineapples in the sample.

The sample proportion of defective pineapples is given by:

p̂ = x/n where x is the number of defective pineapples in the sample and n is the sample size.

p̂ = 65/500 = 0.13

To find the standard error of the estimate, we use the formula:

SE = sqrt(p̂(1-p̂)/n)

SE = sqrt(0.13(1-0.13)/500) = 0.022

We can use the sample proportion and standard error to construct a confidence interval for the true proportion of defective pineapples in the consignment. With a 95% confidence level, the interval is given by:

p̂ ± z*SE

where z is the critical value for the standard normal distribution at a 95% confidence level (z = 1.96)

Therefore, the 95% confidence interval for the true proportion of defective pineapples in the consignment is:

p̂ ± 1.96SE = 0.13 ± 1.960.022 = [0.086, 0.174]

This interval can be converted to a percentage interval by multiplying by 100. So, the interval becomes [8.6%, 17.4%]. We can be 95% confident that the true proportion of defective pineapples in the consignment is between 8.6% and 17.4%.

It's almost certain that the percentage of bad pineapples in the consignment lies between 8.6% to 17.4%.

To know more about the concept please go through the links :

https://brainly.in/question/54161281

https://brainly.in/question/49681614

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