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Probability of sample proportions example

Probability of sample proportions example.

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• How we would solve this if we aren't using a fancy calculator?
• Figure out how many standard deviations away from the mean your proportion is, then consult a z-table and figure out the values.

In other words, since the mean is 0.15 and we want to figure out what the probability that it's greater than 0.10, then the distance from our proportion to the mean is 0.05. Divide this number by the standard deviation to see how many std. dev. it is away from the mean, so 0.05/0.028, and we get 1.77.

We know that this is to the left of the mean, so we're going to use -1.77 when we consult our z-table, which gives us a value of 0.0384. That number is the probability which is BELOW the 0.10 line, so we just subtract that number from 1, and we get approximately 96%.
• But if we know the true proportion to calculate np, we are already know the true proportion why to take samples at all? Its contradicting. As a sampler all I have is sampling data, not true proportions. So how is np threshold a valid approach?
• what happen's when a distribution is not normal?
• In that case, the normal distribution is no more a good approximation of the binomial distribution, and we use the binomial distribution instead.
• How would I do this if I were to standardize the distribution?
• When estimating normality of a sampling distribution do you use the SAMPLE PROPORTION (p̂=0.10) or POPULATION PROPORTION (p=0.15)?

In this case, the surveyors only know that p̂=0.10. And can only estimate normality in that case.
n*p̂=(160)(0.10)=16
n*(1-p̂)=(160)(0.90)=144

Based on the results using p̂, I conclude the sampling distribution is normal. Additionally, using p̂, the cumulative probability of p>0.10 is approximately 50%. Can someone explain to me why the logic of using p̂ is incorrect?
• If you think about it, the sample proportion could be crazily unrepresentative of the actual population proportion. The SRS could have all 160 be really stressed out, and so p-hat would be 1. Obviously, it would be crazy to use p-hat then, since it's so far off. We could instead take a bunch of SRSs, find each p-hat, then take the mean of all these individual sample proportions; this would make it extremely unlikely that p-hat is very far off from the truth. But the mean of a bunch of p-hats is just p, the population proportion!

Honestly, I don't completely get it either so someone else could come along and help!
• Hi, is there a proof of the "expected success and failure number being greater than 10" rule-of-thumb's veracity? Maybe using the Central Limit Theorem or something? It seems very cookbookish and it frustrates me that don't have any deeper intuition for the rule.
• Since this rule was invented by statisticians, it can't really be "proved." It's just that they chose 10 because if the number of successes and failures was less than that, a normal distribution became exceedingly unlikely. But they could have as easily chosen 11 or 12 for the cutoff.
• All problems in this category appear impractical. You can't answer these without knowing the population proportion, but if you knew that, why would you be drawing a random sample to estimate the statistic anyway? All fake examples - impractical
• These examples are designed to teach statistical concepts and may not always represent practical real-life scenarios. They're useful for understanding how to apply statistical methods and for learning to interpret statistical results. In real-world applications, the true population proportion is often unknown, and these methods are used to estimate it from samples.
(1 vote)
• What is the difference the binomial distribution and sampling ditribution? I know that sampling distribution is taking a lot of sample and calculate their statistics, while binomial distribution may only have one sample and the distribution we use is related to population. But I don't know how to differentiate if it is a binomial distribution or sampling distribution from the statement
• Binomial Distribution is used for experiments with two outcomes (success or failure) and is concerned with the number of successes in a fixed number of trials, with each trial being independent and having the same probability of success. If you're dealing with a scenario where you're counting the number of successes in a certain number of trials, and these trials are independent with a fixed probability of success, you're likely looking at a binomial distribution situation.

Sampling Distribution, on the other hand, refers to the distribution of a particular statistic (like the mean or proportion) obtained from a large number of samples drawn from the same population. It is about understanding the behavior of a statistic across different samples from the same population. When a question involves drawing samples and then calculating a statistic (mean, proportion) for each sample to understand the variability of that statistic, it's dealing with a sampling distribution.
(1 vote)
• How can i calculate the probability value without calculator?
(1 vote)
• You probably can't. At the very least you will need a table of the cumulative standard normal probability distribution. There are lots of these on the web. Here, for instance.

https://www.thoughtco.com/standard-normal-distribution-table-3126264

In this example, the population mean is given as .15. Assuming your sample is drawn randomly, this will also be the sample mean. The standard deviation is the square root of (0.15 * 0.85 / 160) ... you'll need a calculator for that, unless you're good at finding square roots with a pencil and paper. That can be done, but it isn't easy. Anyway, the square root mentioned is .0282, very nearly. The difference between the population mean and the observation is 0.15 less 0.10, or 0.05, which is 1.77 standard deviations (.05 / .0282 = 1.77). If you look in the table mentioned above you'll find the Z-value 0.962 listed under 1.77. So there's your answer of 96%.