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The candy machine Suppose a large candy machine has 45\(\%\) orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion \(\hat{p}\) of orange candies. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\) . Check to see if the 10\(\%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Normal condition is met. (d) If the sample size were 50 rather than \(25,\) how would this change the sampling distribution of \(\hat{p} ?\)

Short Answer

Expert verified
(a) Mean is 0.45; (b) SD is 0.0995; condition met; (c) Approx. Normal; (d) Smaller variance.

Step by step solution

01

Calculate the Mean of the Sampling Distribution (a)

The mean of the sampling distribution of \(\hat{p}\) is equal to the true proportion of orange candies in the population. Since 45\% of the candies are orange, the mean is \(\mu_{\hat{p}} = p = 0.45\). This is because the sample proportion is an unbiased estimator of the population proportion.
02

Calculate the Standard Deviation and Check the 10% Condition (b)

The standard deviation of the sampling distribution of \(\hat{p}\) is given by the formula \[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\]. Here, \(p = 0.45\) and \(n = 25\). So, \(\sigma_{\hat{p}} = \sqrt{\frac{0.45(1-0.45)}{25}} \approx 0.0995\). The 10\% condition states that the sample size must be less than 10\% of the population size. If the machine contains many more than 250 candies, the condition is met.
03

Determine if the Sampling Distribution is Approximately Normal (c)

To determine if the sampling distribution is approximately Normal, we check the Normal condition, which requires that both \(np\) and \(n(1-p)\) are greater than 10. Here, \(np = 25 \times 0.45 = 11.25\) and \(n(1-p) = 25 \times 0.55 = 13.75\). Both quantities are greater than 10, so the sampling distribution of \(\hat{p}\) can be approximated by a Normal distribution.
04

Effect of Changing Sample Size to 50 (d)

If the sample size increases to 50, the standard deviation of the sampling distribution changes to \(\sigma_{\hat{p}} = \sqrt{\frac{0.45(1-0.45)}{50}} \approx 0.0707\). The mean remains the same at \(0.45\), but the sampling distribution becomes more concentrated around the mean (tighter) due to the smaller standard deviation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion, often denoted as \( \hat{p} \), is the ratio of elements within a sample that have a specific characteristic. In this case, it's the proportion of orange candies in the sample drawn from the candy machine. The sample proportion serves as an unbiased estimator of the population proportion, meaning on average, it correctly reflects the proportion in the entire population.
The formula for determining \( \hat{p} \) is:
  • \( \hat{p} = \frac{x}{n} \)
- where \( x \) is the count of orange candies in the sample, and \( n \) is the sample size.
This value provides insight into the underlying proportion of the entire population of candies when taking random samples. However, variability exists since each sample may not perfectly mirror the population.
Standard Deviation
The standard deviation of the sampling distribution of \( \hat{p} \) measures the variability of the sample proportion around the population proportion \( p \). It tells us how much we can expect \( \hat{p} \) to vary from sample to sample.
To calculate this, use the formula:
  • \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \]
In our example, with \( p = 0.45 \) and \( n = 25 \), the standard deviation is approximately \( 0.0995 \). This value indicates that the sample proportion will typically vary by about 0.0995 from the true proportion in the population.
Lower standard deviation implies that repeated sampling would yield sample proportions closer to the true population proportion.
Normal Condition
The normal condition is crucial to determine if the sampling distribution of \( \hat{p} \) can be approximated by a normal distribution. This is important because a normal distribution makes it easier to apply statistical inference methods.
To check if the normal condition is satisfied, we ensure:
  • \( np > 10 \)
  • \( n(1-p) > 10 \)
In our case, both calculations:\( np = 25 \times 0.45 = 11.25 \) and \( n(1-p) = 25 \times 0.55 = 13.75 \), are greater than 10, thus confirming that the sampling distribution is approximately normal. This supports analyses that require normality assumptions, such as hypothesis tests and confidence intervals.
10% Condition
The 10% condition ensures that the sample size is sufficiently small relative to the population to allow the model assumptions to hold. This condition helps maintain the independence of trials within the sample.
Specifically, the sample size, \( n \), should be less than 10% of the population size, or \( n < 0.10N \) where \( N \) is the population size.
In this context, if the candy machine contains significantly more than 250 candies, then our sample of 25 satisfies this condition. This means any outcomes from our sample should reliably reflect those possible in the entire population without being affected by the depletion of available options.

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Most popular questions from this chapter

The name for the pattern of values that a statistic takes when we sample repeatedly from the same population is (a) the bias of the statistic. (b) the variability of the statistic. (c) the population distribution. (d) the distribution of sample data. (e) the sampling distribution of the statistic.

A newborn baby has extremely low birth weight \((\mathrm{ELBW})\) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children. Their mean weight at birth was \(\overline{x}=810\) grams. This sample mean is an unbiased estimator of the mean weight \(\mu\) in the population of all ELBW babies, which means that (a) in all possible samples of size 219 from this population, the mean of the values of \(\overline{x}\) will equal 810 . (b) in all possible samples of size 219 from this population, the mean of the values of \(\overline{x}\) will equal \(\mu .\) (c) as we take larger and larger samples from this population, \(\overline{x}\) will get closer and closer to \(\mu\) (d) in all possible samples of size 219 from this population, the values of \(\overline{x}\) will have a distribution that is close to Normal. (e) the person measuring the children's weights does so without any systematic error.

How many people in a car? A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban inter- change. Suppose that this count has mean 1.5 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hours. (a) Could the exact distribution of the count be Normal? Why or why not? (b) Traffic engineers estimate that the capacity of the interchange is 700 cars per hour. Find the probability that 700 cars will carry more than 1075 people. Show your work. (Hint: Restate this event in terms of the mean number of people \(\overline{x}\) per car.)

Do you go to church? The Gallup Poll asked a random sample of 1785 adults whether they attended church or synagogue during the past week. Of the respondents, 44\(\%\) said they did attend. Suppose that 40\(\%\) of the adult population actually went to church or synagogue last week. Let \(\hat{p}\) be the proportion of people in the sample who attended church or synagogue. (a) What is the mean of the sampling distribution of \(\hat{p}\) ? Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p} .\) Check to see if the 10\(\%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Normal condition is met. (d) Find the probability of obtaining a sample of 1785 adults in which 44\(\%\) or more say they attended church or synagogue last week. Do you have any doubts about the result of this poll?

Airport security The Transportation Security Administration \((\mathrm{TSA})\) is responsible for airport safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers \(-12\) in first class and 64 in coach class. TSA officers selected an SRS of 10 passengers for screening. Let \(\hat{p}\) be the proportion of first-class passengers in the sample. (a) Is the 10\(\%\) condition met in this case? Justify your answer. (b) Is the Normal condition met in this case? Justify your answer.

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