Use the result of part (c) of Exercise 58 to find the probability that in a random sample of siz

Use the result of part (c) of Exercise 58 to find the...

Key Concepts

Nonparametric Inference

Nonparametric inference involves drawing conclusions from data without making strong assumptions about the underlying probability distribution. Using the sample extremes (order statistics) to predict population coverage is a typical nonparametric method, emphasizing distribution-free techniques that rely on ranks or order rather than specific distribution parameters.

Binomial Probability

Binomial probability is used when evaluating the likelihood of a certain number of successes in a fixed number of independent trials. In the context of determining the probability that a sample interval covers a specified proportion of the population, combinatorial and binomial arguments can be employed to assess the chances that additional observations lie within the calculated range.

Order Statistics

Order statistics refer to the sorted values obtained from a sample. In this context, the smallest and largest observations form the endpoints of an interval that can be used to estimate or capture a certain percentage of the population. They are fundamental in nonparametric analyses as they do not depend on underlying distributional assumptions.

Tolerance Intervals

Tolerance intervals are statistical intervals computed from a sample that are designed to contain at least a specified proportion of the population with a given level of confidence. The problem uses the sample extremes to construct an interval that is expected to contain at least 80% of the population, a concept widely applied in quality control and process management.

Transcript

00:01 So this problem is a binomial distribution problem and it tells us that a simple random sample of 75 is obtained from a population with a proportion with a specified characteristic of p equals 0 .8.

00:14 Let's start with part a, which asks us to describe the sampling distribution of p hat.

00:19 Before we do that, we'll have to define our three important variables of n, p, and q.

00:25 So it gives us n, which is 75, that's the sample size.

00:29 It gives us p as well, which is 0 .8.

00:33 And then q is simply 1 minus p, so 1 minus 0 .8 is just 0 .2.

00:39 Now with these variables, we're ready to find our mean and standard deviation, as i have the equations for all the screen.

00:46 So starting with the mean, the mean is equal to n times p, so 75 times 0 .8, which comes out to 60.

00:54 Standard deviation is equal to the square root of n times p, so that's the square root of 7.

01:00 Or excuse me the square root of n times p times q so it's a square root of 75 times 0 .8 times 0 .2 and this is equal to 3 .464.

01:14 So our since our mean is greater than 5 that means that we are dealing with a normal distribution, mean of 60 and a standard deviation of 3 .464.

01:31 Now we'll move on to part b which asks us to define the the probability of obtaining 63 or more individuals with this characteristic.

01:41 So what we're finding is the probability of x being greater than or equal to 63 since it asks for 63 or more.

01:49 Now because of this equal sign here we'll have to do a continuity correction for 63, meaning we'll have to subtract 0 .5.

01:56 So what we're really finding is the probability of x being greater than 62 .5...

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