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Sample Proportions Section 9.2

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1 Sample Proportions Section 9.2
Which of the following best describes the sampling distribution of a mean? It is the particular distribution in which πœ‡ π‘₯ =πœ‡ and 𝜎 π‘₯ =𝜎. It is a graphical representation of the means of all possible samples. It is the distribution of all possible sample means from a given population. It is the distribution of all possible sample means of a given size. It is the probability distribution for each possible sample size. Sample Proportions Section 9.2

2 Recall If a random variable 𝑿 has a distribution 𝑩(𝒏, 𝒑) then: πœ‡ 𝑋 =𝑛𝑝
(This is on your formula packet!) If a random variable 𝑿 has a distribution 𝑩(𝒏, 𝒑) then: πœ‡ 𝑋 =𝑛𝑝 𝜎 𝑋 = 𝑛𝑝 1βˆ’π‘

3 Also recall For 𝑿: πœ‡ 𝑋 and 𝜎 𝑋 If 𝒂𝑿: πœ‡ π‘Žπ‘‹ =π‘Ž πœ‡ 𝑋 , and 𝜎 π‘Žπ‘‹ =π‘Ž 𝜎 𝑋
Multiplying each term in a distribution by the same constant has the effect of multiplying both the mean and standard deviation of that distribution by that constant. For 𝑿: πœ‡ 𝑋 and 𝜎 𝑋 If 𝒂𝑿: πœ‡ π‘Žπ‘‹ =π‘Ž πœ‡ 𝑋 , and 𝜎 π‘Žπ‘‹ =π‘Ž 𝜎 𝑋 (multiplying)

4 The Sampling Distribution of 𝒑
Sampling Distribution – distribution of all possible samples of same size and from same population. The mean of the sampling distribution of 𝒑 is exactly 𝒑. πœ‡ 𝑝 =𝑝 The standard deviation of the sampling distribution of 𝑝 is: 𝜎 𝑝 = 𝑝 1βˆ’π‘ 𝑛 Rule of Thumb #1: Use this formula for standard deviation of 𝑝 only when the population is at least 10 times as large as the sample. (𝑁β‰₯10𝑛)

5 Examining Standard Deviation of 𝒑
𝜎 𝑝 = 𝑝 1βˆ’π‘ 𝑛 Since 𝒏 is in the denominator, what will happen to 𝜎 𝑝 when 𝒏 increases? 𝜎 𝑝 will decrease ( 𝜎 𝑝 is less variable in larger samples) By what factor would we need to increase n in order to decrease 𝜎 𝑝 by 1 2 ? 4, since = 1 2

6 Using the Normal Approximation for 𝒑
The Binomial distribution has a an approximately Normal shape, but only when sample sizes are large Rule of Thumb #2 We will use the Normal approximation to the sampling distribution of 𝑝 for values of 𝑛 and 𝑝 that satisfy: 𝑛𝑝β‰₯10 and 𝑛 1βˆ’π‘ β‰₯10

7 Example – Applying to College
A polling organization asks an SRS of 1500 first-year college students whether they applied for admission to any other college. We know that 35% of all first-year college students applied to colleges besides the one they are attending. What is the probability that the random sample of 1500 students will give a result within 2 percentage points of this true value? How do you think we will solve this problem? Don’t forget to pay attention to the Rules of Thumb! We will use Normal approximation.

8 Applying to college (cont)
We know that 𝑛 = 1500, and 𝑝 = 0.35 The sampling distribution of 𝑝 has πœ‡ 𝑝 =0.35 In order to use our formula for standard deviation the population must be at least 10 times the size of the sample. Check Rule of Thumb 1 So the population must be at least 10βˆ™1500=15,000 people. There are over 1.7 million first year college students, so we’re ok 𝜎 𝑝 = 𝑝 1βˆ’π‘ 𝑛 = 𝜎 𝑝 =0.0123

9 Applying to college (cont)
Can we use a Normal distribution to approximate the sampling distribution of 𝑝 ? Rule of Thumb 2 𝑛𝑝= =525 𝑛 1βˆ’π‘ = =975 Yes, we can.

10 Normal Approximation Since we want to find the probability that 𝑝 falls within 2 percentage points of 0.35, what values do we want to calculate? We want to find the probability that 𝑝 falls between 0.33 and 0.37 0.33≀ 𝑝 ≀0.37 𝑧= 𝑝 βˆ’ where 𝑝 =0.33 and 𝑝 =0.37 𝑧=βˆ’1.63 and 1.63 𝑃 0.33≀ 𝑝 ≀0.37 =𝑃 βˆ’1.63≀𝑧≀ SKETCH A NORMAL CURVE! =0.9484βˆ’0.0516=0.8968 About 90% of all samples will give a result within 2 percentage points of the truth about the population.


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