b:head_first_statistics:estimating_populations_and_samples
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| b:head_first_statistics:estimating_populations_and_samples [2024/11/11 08:08] – [Recap] hkimscil | b:head_first_statistics:estimating_populations_and_samples [2025/10/08 12:20] (current) – [Sampling distribution of sample mean] hkimscil | ||
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| $\hat\mu$ : See this hat I’m wearing? It means I’m a point estimator. If you don’t have the exact value of the mean, then I'm the next best thing. | $\hat\mu$ : See this hat I’m wearing? It means I’m a point estimator. If you don’t have the exact value of the mean, then I'm the next best thing. | ||
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| p = 32/40 = 0.8 | p = 32/40 = 0.8 | ||
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| Mighty Gumball takes another sample of their super-long-lasting gumballs, and finds that in the sample, 10 out of 40 people prefer the pink gumballs to all other colors. What proportion of people prefer pink gumballs in the population? What’s the probability of choosing someone from the population who doesn’t prefer pink gumballs? | Mighty Gumball takes another sample of their super-long-lasting gumballs, and finds that in the sample, 10 out of 40 people prefer the pink gumballs to all other colors. What proportion of people prefer pink gumballs in the population? What’s the probability of choosing someone from the population who doesn’t prefer pink gumballs? | ||
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| ===== Exercise ===== | ===== Exercise ===== | ||
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| 25% of the gumball population are red. What’s the probability that in a box of 100 gumballs, at least 40% will be red? We’ll guide you through the steps. | 25% of the gumball population are red. What’s the probability that in a box of 100 gumballs, at least 40% will be red? We’ll guide you through the steps. | ||
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| ====== Sampling distribution of sample mean ====== | ====== Sampling distribution of sample mean ====== | ||
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| According to Mighty Gumball’s statistics for the population, the mean number of gumballs in each packet is 10, and the variance is 1. The trouble is they’ve had a complaint. One of their most faithful customers bought 30 packets of gumballs, and he found that the average number of gumballs per packet in his sample is only 8.5. | According to Mighty Gumball’s statistics for the population, the mean number of gumballs in each packet is 10, and the variance is 1. The trouble is they’ve had a complaint. One of their most faithful customers bought 30 packets of gumballs, and he found that the average number of gumballs per packet in his sample is only 8.5. | ||
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| ===== Exercise ===== | ===== Exercise ===== | ||
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| Let’s apply this to Mighty Gumball’s problem. | Let’s apply this to Mighty Gumball’s problem. | ||
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| ====== Recap ====== | ====== Recap ====== | ||
| - | Distribution of **Sample** <fc # | + | Distribution of **Sample** <fc # |
| + | when sampling n entities (repeatedly) from a population whose proportion is p. | ||
| + | \begin{eqnarray*} | ||
| + | Ps & \sim & N(p, \frac{pq}{n}) \\ | ||
| + | \text{hence, | ||
| + | \text{standard deviation of} \\ | ||
| + | \text{sample proportions} & = & \sqrt{\frac{pq}{n}} | ||
| + | \end{eqnarray*} | ||
| Distribution of **Sample** <fc # | Distribution of **Sample** <fc # | ||
| + | when sampling a sample whose size is n from a population whose mean is $\mu$ and variance is $\sigma^2$. | ||
| + | \begin{eqnarray*} | ||
| + | \overline{X} & \sim & N(\mu, | ||
| + | \text{hence, | ||
| + | \text{standard deviation of} \\ | ||
| + | \text{sample means} & = & \sqrt{\frac{\sigma^2}{n}} \\ | ||
| + | & = & \frac{\sigma}{\sqrt{n}} | ||
| + | \end{eqnarray*} | ||
b/head_first_statistics/estimating_populations_and_samples.1731280084.txt.gz · Last modified: by hkimscil