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Sampling Distribution Using the Z-table. Review: on using TI-84 Find the area of the following normal distribution using the Z table 1. z > 2.85 2. z.

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Presentation on theme: "Sampling Distribution Using the Z-table. Review: on using TI-84 Find the area of the following normal distribution using the Z table 1. z > 2.85 2. z."— Presentation transcript:

1 Sampling Distribution Using the Z-table

2 Review: on using TI-84 Find the area of the following normal distribution using the Z table 1. z > 2.85 2. z < 2.85 3. z > -1.66 4. -1.66< z <2.85 Syntax: 2nd>vars>normalcdf>min,max,0,1) +infinity=1EE99 -infinity=-1EE99 0.0022 0.9978 0.9515 0.9493

3 Review: on using TI-84 Find the area of the following normal distribution using the Z table Syntax: 2nd>vars>normalcdf>min,max,mean,sd) +infinity=1EE99 -infinity=-1EE99 0.2266 0.2260 0.6915 1. x > 0.40, mean: 0.37, sd: 0.04 2. 0.40 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1423908/4/slides/slide_2.jpg", "name": "Review: on using TI-84 Find the area of the following normal distribution using the Z table Syntax: 2nd>vars>normalcdf>min,max,mean,sd) +infinity=1EE99 -infinity=-1EE99 0.2266 0.2260 0.6915 1.", "description": "x > 0.40, mean: 0.37, sd: 0.04 2. 0.40

4 Sampling distribution of sample proportion p Count of success in sample Size of the sample X n == The mean of the sampling distribution is exactly p p The standard deviation of the sampling distribution is p p(1-p) n

5 One way of checking the effect of under-coverage, non response, and other sources of error in a sample survey is to compare the sample with the known facts about the population. ƥ =0.11 =0.11 σ=0.00808 p(1-p) n σ = About 11% of American adults are black, the proportion of blacks in an SRS of 1500 adults should therefore be close to 0.11. It is unlikely to be exactly 0.11 because of sampling variability. If a national sample contains only 9.2% blacks should we suspect the sampling procedure is somehow under-representing blacks? X=.092

6 probability ƥ =0.092 P( ƥ 0.092) = 0.0129 Only 1.29% of all samples would have so few blacks. Therefore we have a good reason to suspect that the sampling procedure underrepresented blacks.

7 Rule of Thumb 1. You can only use the formula for the standard deviation of p-hat only when the population is at least 10 times as large as the sample N 10n 2. Use the normal approximation to the sampling distribution of p-hat for values of n and p that satisfy np 10 and n(1-p) 10

8 Practice: Rule of thumb(s) Explain why you cannot use the methods in ch9.1 on this problem A factory employs 3000 unionized workers, of whom 30% are hispanic. The 15-member union executive committee contains 3 hispanics. What would be the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the worker. N 10n 3000 10(15) 3000 150 satisfied np 10 and n(1-p) 10 15(.30) 10 and 15(1-30) 10 4.5 10 and 10.5 10 NOT satisfied

9 Mean and Standard Deviation of a sample mean Mean of sampling distribution: x = Standard Deviation of sampling distribution: σ x = σ / n

10 Mean and Standard Deviation of a sample mean Investors remember 1987 as the year the stocks lost 20% of their value in a single day. For 1987 as a whole, the mean return of all common stocks on the NYSE was =- 3.75% and the standard deviation of the returns was about σ=26%. What are the mean and standard deviation of the distribution for all possible samples of 5 stocks? = -3.75% or -.00375 σ x = σ / n σ= 26 / 5 σ= 11.6376%

11 Example: Servicing Air conditioners The average of servicing an air conditioning unit in a certain company is 1 hour with a standard deviation of 1 hour as well. The company has been contracted to maintain 70 of these units in an apartment building. You must schedule technicians time for a visit to this building. Is it safe to budget an average of 1.1 hours for each unit? Or should you budget 1.25 hours? =1, σ=1 Sd=σ/n 1/70=0.120hrs

12 =1 hr. σ=.120 hr. N(1,.120) normalcdf(1.1, +, 1,.120) P(x-bar>1.1 hrs.)P(x-bar>1.25 hrs.) = 0.2033 normalcdf(1.25, +, 1,.120) = 0.0182 If you budget 1.1 hrs, there is a 20% chance that the technicians will not be able to complete the work If you budget 1.25 hrs, there is a 2% chance that the technicians will not be able to complete the work

13 What you should have learned 1. Identify parameters and statistics in sample experiment. 2. Recognize the fact of sampling variability. 3. Interpreting sampling distribution. 4. Describe the bias and variability of statistic in terms of the mean and its spread. 5. Understand the variability of a statistic. Statistic from larger samples are less variable. A. Sampling Distribution

14 1. Recognize when a problem involves a sample proportion ƥ 2. Find the mean and standard deviation of sampling distribution 3. Know that as the spread gets smaller the sample size gets bigger. 4. Recognize a reliable conclusion by verifying the rule of thumbs: N10n and np10, nq10 B. Sample Proportions

15 1. Recognize when a problem involves the mean of the sample. (x-bar) 2. Find the mean and sd of the sampling distribution when the and σ of the population are known. 3. Know that as the spread gets smaller the sample size gets bigger 4. X-bar is approximately Normal when the sample size is large (CLT) C. Sample Means


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