Confirm that the sample is large enough to assume that the sample proportion is normally distributed.Compute the sample proportion of items shipped within 12 hours.A consumer group placed 121 orders of different sizes and at different times of day 102 orders were shipped within 12 hours. To be within 5 percentage points of the true population proportion 0.38 means to be between 0.38 − 0.05 = 0.33 and 0.38 + 0.05 = 0.43. Which lies wholly within the interval, so it is safe to assume that P ^ is approximately normally distributed. First we use the formulas to compute the mean and standard deviation of P ^: μ P ^ = p = 0.38 and σ P ^ = p q n = ( 0.38 ) ( 0.62 ) 900 = 0.01618 The information given is that p = 0.38, hence q = 1 − p = 0.62. Find the probability that the sample proportion computed from a sample of size 900 will be within 5 percentage points of the true population proportion.Verify that the sample proportion P ^ computed from samples of size 900 meets the condition that its sampling distribution be approximately normal.Nine hundred randomly selected voters are asked if they favor the bond issue. Suppose that in a population of voters in a certain region 38% are in favor of particular bond issue. However, the condition that the sample be large is a little more complicated than just being of size at least 30. Thus the Central Limit Theorem applies to P ^. In the same way the sample proportion p ^ is the same as the sample mean x. Thus the population proportion p is the same as the mean μ of the corresponding population of zeros and ones. Clearly the proportion of the population with the special characteristic is the proportion of the numerical population that are ones in symbols, p = number of 1 s Nīut of course the sum of all the zeros and ones is simply the number of ones, so the mean μ of the numerical population is μ = Σ x N = number of 1 s N This gives a numerical population consisting entirely of zeros and ones. To see how, imagine that every element of the population that has the characteristic of interest is labeled with a 1, and that every element that does not is labeled with a 0. The Central Limit Theorem has an analogue for the population proportion P ^. The mean μ P ^ and standard deviation σ P ^ of the sample proportion P ^ satisfy μ P ^ = p and σ P ^ = p q n Suppose random samples of size n are drawn from a population in which the proportion with a characteristic of interest is p. μ P ^ and a standard deviation A measure of the variability of proportions computed from samples of the same size. It has a mean The number about which proportions computed from samples of the same size center. Viewed as a random variable it will be written P ^. The sample proportion is a random variable: it varies from sample to sample in a way that cannot be predicted with certainty. Thus if in reality 43% of people entering a store make a purchase before leaving, p = 0.43 if in a sample of 200 people entering the store, 78 make a purchase, p ^ = 78 / 200 = 0.39. The population proportion is denoted p and the sample proportion is denoted p ^.
Often sampling is done in order to estimate the proportion of a population that has a specific characteristic, such as the proportion of all items coming off an assembly line that are defective or the proportion of all people entering a retail store who make a purchase before leaving. To learn what the sampling distribution of P ^ is when the sample size is large.To understand the meaning of the formulas for the mean and standard deviation of the sample proportion.To recognize that the sample proportion P ^ is a random variable.I am not sure, why when computing SE, we don't use p, but use $\bar Y$ instead. Then the methods to compute SE should be using the formula above. I know it is Bernoulli, but I don't know its real p(probability of success). However, i am considering the case where i know i am drawing samples from a Bernoulli distribution this has two cases. The formula should be SE = sample standard deviation / sqrt(n). This should be based on the case where the population std is unknown. I know that standard error of the sample average Y_bar should be an estimator of the standard deviation of the sampling distribution $\bar Y$. I am confused about the formula here about standard error.