Alta - Chapter 7 - The Central Limit Theorem - Part 2
Sampling Distribution for Sample Proportions
As we were for sample means, we are interested in the distribution of possible sample proportions for a given sample size n. This sampling distribution for the sample proportion p^ has mean μp^ and standard deviation σp^. μp^, the mean of the sampling distribution of sample proportions, shares the same mean with the population distribution. That is, μp^=p, the population proportion. σp^, the standard deviation of the sampling distribution of sample proportions, depends on both the population proportion p and the sample size n, and is given by the formula σp^=p(1−p)n−−−−−√. The relationship between the two distributions is summarized below.
Example of Proportion in Statistics
The researcher is interested in proving that people on parole are more likely to be re-convicted of a second crime than not. In a study of 50 inmates who were released from prison on parole, the researcher found that 30 of them had been re-convicted of a second crime within 2 weeks of being released. In this example, there are n=50 trials, of which x=30 are successes. Thus, we say that the population proportion is p=3050=0.6. Let's say the same researcher chose 30 out of the original 50 inmates to follow and found that, of those 30, 25 were convicted of a second crime within two weeks of being released from prison. In this example, there are 30 inmates in the sample, and 25 "successes." The sample proportion is p^=2530=0.83. Note: A "Success" is not always a good thing. A "Success" is just the characteristic of interest.
A youth director for a city's parks and recreations department is planning for next year's program. From a survey of city residents, the director knows that 35% of the population participate in the parks and recreations programs. For a random sample of 40 people, what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?
Correct answers:$0.075$0.075 Given the population proportion p=35%=0.35 and a sample size of n=40, the standard deviation of the sampling distribution of sample proportions is σpˆ=p(1−p)n−−−−−−−√=0.35(1−0.35)40−−−−−−−−−−−−√≈0.075
From recent survey data, its known that the proportion of adults in the United States who are smokers is 18%. For a random sample of size 150, what is the standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?
Correct answers:$0.031$0.031 Given the population proportion p=18%=0.18 and a sample size of n=150, the standard deviation of the sampling distribution of sample proportions is σpˆ=p(1−p)n−−−−−−−√=0.18(1−0.18)150−−−−−−−−−−−−√≈0.031
At a small business convention, 63% of businesses turned a profit within the first three years of opening. A reporter would like to interview these successful businesses. If a random sample of 60 businesses at the convention are chosen, Use the graph below to calculate probability that more than 80% of them turned a profit within the first three years of opening. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to three decimal places.
1. We will select the greater than graph option, with the purple area in the right tail of the curve, because we want to find P(p^>0.80), the probability that the sample proportion p^ will be more than 80%=0.80. We are given a population proportion of p=0.63. We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10 andnp≥10. We have a sample size of 60 businesses and p=0.63, so we have: 60(1−0.63)=22.2 and 60(0.63)=37.8. 22.2 and 37.8 are greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We know that μp^=p=0.63 by the Central Limit Theorem; Substituting p=0.63 and n=60, we find the standard error to beσp^=0.63(1−0.63)60−−−−−−−−−−−−√=0.0623 We are given p^=0.80. Then the z-score corresponding to p^=80%=0.80 is z=0.80−0.630.06=2.83 3. Slide the black dot to the z-score we found above: 2.83. The area shown is 0.0023. Rounding to three decimal places, we have P(p^>0.80)=0.002. So, the probability that more than 80% of a random sample turned a profit within the first three years of opening is 0.002, or 0.2%.
Population & Sample Proportions
A Population Proportion, p=\dfrac{x}{n}p=xn, is the number of successes in the population divided by the total population size. A Sample Proportion, \hat{p}=\frac{x}{n}p^=xn, is the number of successes in a sample divided by the sample size, or number of trials nn . It is important to remember that \hat{p}p^ is an estimate of the population proportion. Sometimes it is referred to as a point estimate.
To fairly divide their patients on a hospital ward, oncology nurses are going to examine their typical sample proportions. They find that, of 780 patients admitted to a hospital at any given time, there are 130 patients on the oncology ward. In the sampling distribution of sample proportions of size 36 patients, above what proportion will 75% of all sample proportions be? Select all answers that apply to your calculation below. Use the z-table given below to answer the question:
Correct answer: z=−0.68 p^=0.13 Since we are looking for the top 75% of sample proportions, we need to look in the z-table for the bottom 25%. So the top 75% corresponds to a z-score of z=−0.68. Next, we will find the sample proportion that goes with the z-score. Note the population proportion is not directly given, but can be calculated as follows: p=130780=0.167. To find the sample proportion p^ that corresponds to z=−0.68, we substitute the known values into the formula z=p^−pσp^ and solve for p^ to get −0.68p^=p^−0.1670.167(1−0.167)36−−−−−−−−−√≈0.125 Rounded to two decimal places, p^=0.13. This means: the top 75% of possible sample proportions are at or above 0.13.
From recent census data, it is discovered that the proportion of the adults in the United States who are first generation Americans is 14%. For a random sample of size 500, what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?
Correct answers:$0.016$0.016 Given the population proportion p=14%=0.14 and a sample size of n=500, the standard deviation of the sampling distribution of sample proportions is σpˆ=p(1−p)n−−−−−−−√=0.14(1−0.14)500−−−−−−−−−−−−√≈0.016
A college financial adviser is interested in studying the number of students at his college who apply for loans when entering graduate school. Of the 721 students at the college, 299 of them apply for loans when entering graduate school. If the financial adviser conducts a study and samples 45 students at the college who are starting graduate school next fall, what is the probability that more than 16 of them apply for loans? Round your answer to two decimal places. Use this z-table.
Correct answers:$0.80\pm0.01$0.80±0.01 We are given a population proportion of p=0.415. For samples of size n=45, we want to find P(p^>0.356), the probability that the sample proportion p^ will be more than 35.6%=0.356. The Central Limit Theorem for sample proportions tell us that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. Substituting μp^=0.415 and n=45, we find the standard error to be σp^=0.415(1−0.415)45−−−−−−−−−−−−−−√=0.073 Since p^ is normally distributed, we can compute the probability using the z-score and a standard normal table. For a population proportion of p=0.415, the z-score corresponding to p^=35.6%=0.356 is z=0.356−0.4150.073=−0.81 From the table, P(p^<0.356)=0.209. This is the probability that p^ is less than 35.6%. Taking the complement, we have P(p^>0.356)=1−0.209=0.791 So, to two decimal places, the probability that more than 35.6% of a random sample will have applied for a loan is 0.80, or 80%.
In a large town, 57% of the population over the age of 25 have a college degree. Use the graph below to calculate the probability that if a random sample of 50 people over the age of 25 are chosen, over half of them have a college degree. Use the graph below to calculate the probability that at least half the people in the sampling distribution will have a college degree: Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Enter your answer in decimal form and round to two decimal places.
Correct answers:$0.84\pm0.01$0.84±0.01 1. We will select the greater than graph option, with the purple area in the right tail of the curve, because we want to find P(p^>0.50), the probability that the sample proportion p^ will be more than 50%=0.50. We are given a population proportion of p=0.57. Next, we will check that the Central Limit Theorem for Sample Proportions applies. n=50 and p=0.57, so n(1−p)=21.5. This is greater than 10, so we may proceed and use the standard curve graph above to find our probability. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^. The Central Limit Theorem for sample proportions tell us that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. We have μp^=p=0.57; The standard deviation of the sampling distribution of sample proportions is: σp^=0.57(1−0.57)50−−−−−−−−−−−−√=0.0700 The sample proportion is p^=0.50. Then, the corresponding z-score is z=0.50−0.570.07=−1 3. Slide the black dots to the z-score we found above: −1. The area shown is 0.8413. Rounding to the nearest hundredth, we have P(p^>50)=0.84.
Pharmacy technicians are concerned about the rising number of fraudulent prescriptions they are seeing. A small pharmacy sees 1,500 new prescriptions a month, 28 of which are fraudulent. Find the population proportion, as well as the mean and standard deviation of the sampling distribution for samples of size n=180. Round all answers to 3 decimal places.
Correct answers:1$0.019$0.0192$0.019$0.0193$0.010$0.010 The population proportion is p=281,500≈0.019. This is also the mean of the sampling distribution: μp^=p=0.019 For samples of size n=180, the standard deviation of the sampling distribution is σpˆ=p(1−p)n−−−−−−−√=0.019(1−0.019)180−−−−−−−−−−−−−−√≈0.010
A baseball team calls itself "America's Favorite Team," because it has 90,000 fans on social media out of 2,210,000 social media users. Find the population proportion, as well as the mean and standard deviation of the sampling distribution for samples of size n=1,000. Round all answers to 3 decimal places.
Correct answers:1$0.041$0.0412$0.041$0.0413$0.006$0.006 The population proportion is p=90,0002,210,000≈0.041. This is also the mean of the sampling distribution: μp^=p=0.041 For samples of size n=1,000, the standard deviation of the sampling distribution is σpˆ=p(1−p)n−−−−−−−√=0.041(1−0.041)1,000−−−−−−−−−−−−−−√≈0.006
According to a study, 40% of people have more than $10,000 in credit card debt. Suppose that in a particular bank there are currently 60 customers waiting in line. Use the graph below to calculate the probability that of those 60 customers, between 21 and 23 of them have more than $10,000 in credit card debt. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to two decimal places.
Correct answers:1$0.06\pm0.01$0.06±0.012$0.17\pm0.01$0.17±0.01 1. We will select the between graph option, with the purple area in the middle of the curve, because we want to know the probability that between 21 and 23 of the sample have credit card debt. We are given the population proportion of p=40%=0.4. We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10 and np≥10. We have a sample size of 60 customers and p=0.4, so we have: 60(1−0.4)=36 and 60(0.4)=24. 36 and 24 are greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We have μp^=p=0.4; The standard deviation of the sampling distribution of sample proportions is thusσp^=0.4(1−0.4)60−−−−−−−−−−√≈0.06 We calculate two p^ values: p^1=2160=0.35 and p^2=2360=0.38. Then the corresponding z scores are: z1=0.35−0.40.06=−0.83andz2=0.38−0.40.06=−0.33 3. Slide the black dots to the two z-scores we found above: −.83 and −.33. The area shown is 0.1674. Rounding to two decimal places, we have P(21≤X≤23)=0.17. That is, of the 60 customers waiting in line, the probability that between 21 and 23 of them have more than $10,000 in credit card debt is 0.17.
In a large city, 70% of people own a cell phone. A sociologist wishes to study the impact cell phones have on society. To get started, the sociologist randomly samples 45 people in the city. Use the graph below to calculate the probability that of those 45 people sampled, between 33 and 36 of them own a cell phone. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to two decimal places.
Correct answers:1$0.07\pm0.01$0.07±0.012$0.26\pm0.01$0.26±0.01 1. We will select the between graph option, with the purple area in the middle of the curve, because we want to know the probability that between 33 and 36 people in the sample own a cell phone. We are given the population proportion of p=70%=0.7. We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10 and np≥10. We have a sample size of 45 people and p=0.7, so we have: 45(1−0.7)=13.5 and 45(0.7)=31.5. 31.5 and 13.5 are greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We know that μp^=p=0.7; The standard deviation of the sampling distribution of sample proportions is thusσp^=0.7(1−0.7)45−−−−−−−−−−√≈0.07 We calculate two p^ values: p^1=3345=0.73 and p^2=3645=0.80. Then, we can calculate the corresponding z-scores: z1=0.73−0.70.07=0.43andz2=0.80−0.70.07=1.43 3. Slide the black dots to the z-scores we found above: 0.43 and 1.43. The area shown is 0.2572. Rounding to two decimal places, we have P(33≤X≤36)=0.26. That is, the probability that between 33 and 36 people own a cell phone out of a sample of 45 people is 0.26.
According to a study, 60% of people who are murdered knew their murderer. Suppose that in a particular state there are currently 50 current cold cases. What is the probability that of those 50 cold cases, between 28 and 33 of them knew their murderer? Round all numbers to two decimal places. Use the z-tables below:
Correct answers:1$0.07\pm0.01$0.07±0.012$0.52\pm0.01$0.52±0.01 We are given the population proportion of p=60%=0.6. Again, we will check the conditions of the CLT. n=50, and p=0.6, so n(1−p)=20. 20 is greater than 10, so we may proceed as if our distribution is normal by using the standard normal table for probabilities The standard deviation of the sampling distribution of sample proportions is thus σp^=0.6(1−0.6)50−−−−−−−−−−√≈0.07 The proportions of interest are then p^1=2850=0.56 and p^2=3350=0.66, which correspond to the two z-scores z1=0.56−0.60.07=−0.57andz2=0.66−0.60.07=0.86 The probabilities corresponding to these z-scores can be found in the given tables: From the z-tables, P(X≤28)=P(z≤−0.57)=0.284 and P(X≤33)=P(z≤0.86)=0.805. The probability that, in a sample of n=50, the sample proportion p^ falls between 2850 and 3350 is the difference between these probabilities: P(28≤X≤33)=P(−0.57≤z≤0.86)=0.805−0.284=0.521 So to two decimal places, P(28≤X≤33)=0.52.
According to research about tech company growth, 67% of successful tech companies spend a majority of their advertising online. Suppose that in a particular state there are currently 32 tech companies. What is the probability that of those 32 tech companies, between 23 and 25 of them spend a majority of their advertising online? Round all numbers to two decimal places. Use the z-tables below:
Correct answers:1$0.08\pm0.01$0.08±0.012$0.18\pm0.01$0.18±0.01 We are given the population proportion of p=67%=0.67. The standard deviation of the sampling distribution of sample proportions is thus σp^=0.67(1−0.67)32−−−−−−−−−−−−√≈0.08 The proportions of interest are then p^1=2332=0.72 and p^2=2532=0.78, which correspond to the two z-scores z1=0.72−0.670.08=0.63andz2=0.78−0.670.08=1.38 The probabilities corresponding to these z-scores can be found in the given tables: From the z-tables, P(X≤23)=P(z≤0.63)=0.736 and P(X≤25)=P(z≤1.38)=0.916. The probability that, in a sample of n=32, the sample proportion p^ falls between 2332 and 2532 is the difference between these probabilities: P(23≤X≤25)=P(0.63≤z≤1.38)=0.916−0.736=0.180 So to two decimal places, P(23≤X≤25)=0.18.
In a town, a pediatric nurse is concerned about the number of children who have whooping cough during the winter season. Of the 3,492 children living in a town, 623 of them have whooping cough. Determine the probability that in sample of 60 children, 7 or fewer of them have whooping cough. Round the z-score to two decimal places and all other answers to three decimal places. Use the z-table below:
Correct answers:1$0.178\pm0.001$0.178±0.0012$0.117\pm0.001$0.117±0.0013$0.049\pm0.001$0.049±0.0014$-1.24\pm0.001$−1.24±0.0015$0.107\pm0.001$0.107±0.001 We know the total number of children in the total population, so we know the population proportion: p=6233,492=0.178 The standard deviation of the sampling distribution of sample proportions is thus σp^=0.178(1−0.178)60−−−−−−−−−−−−−−√=0.049 For the given sample of 60 children, the sample proportion is p^=760=0.117 . The corresponding z-score is z=p^−μp^σp^=0.117−0.1780.049=−1.24 Since we are looking for the probability that the number of children who have whooping cough is less than or equal to 7, we want to know the area under the left tail of the standard normal distribution, to the left of z=−1.24. Using the z-table, z=−1.24 corresponds to a left-tail probability of 0.107, so P(z≤0.117)=0.107. That is, the probability that 7 or fewer out of 60 children have whooping cough is 0.107, when rounded to two decimal places.
Dentists recommend that people get their teeth cleaned at least once a year as a preventative measure to tooth decay and gum disease. A recent study by a dental insurance company shows that 54% of all people visit their dentists yearly. A dentist takes a simple random sample of 49 people. With a 85% chance, the dentist's sample proportion will be no greater than what value of p^? Round your answer to the nearest hundredth. Use the z-table given below:
Correct answers:1$1.03$1.032$0.07$0.073$0.61$0.61 The question asks about the likelihood of sample proportions, so it is the sampling distribution of sample proportions we are concerned with. Since the population proportion is given to be 0.54, we know the mean of the sampling distribution is also 0.54. The standard deviation of the sampling distribution for samples of size 49 is σp^=p(1−p)n−−−−−−−√=0.54(1−0.54)49−−−−−−−−−−−−√≈0.07 We want to know the point in the sampling distribution for which 85% of sample proportions fall to the left and the remaining 15% fall to the right. We can work backwards in the z-table below to find the z-score corresponding to this separating percentage of 85%. We see that 85% corresponds to a z-score of about 1.03. To find the sample proportion p^ that corresponds to this z-score, we substitute the known values into the formula z=p^−pσp^ to get 1.03p^=≈p^−0.540.070.613 If the area under the sampling distribution is 0.85, the value of the sample proportion p^ that separates this area from the other 0.15 units is approximately 0.61 units, when rounded to the nearest hundredth. This means: there is a 85% chance that the dentist's sample proportion will be at or below 0.61.
A college has a total enrollment of 2445 students, and 469 of them are left-handed. Use the graph below to determine the probability that a survey of 50 students will find that 9 or fewer students are left-handed. Use the graph below to calculate the probability that the sampling distribution will have a proportion of left-handed students less than or equal to 9: Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round to two decimal places.
Correct answers:1$0.19\pm0.01$0.19±0.012$0.18\pm0.01$0.18±0.013$0.06\pm0.01$0.06±0.014$-0.17\pm0.01$−0.17±0.015$0.43\pm0.01$0.43±0.01 1. We will select the less than graph option, with the purple area in the left tail of the curve, because we want to know the probability that 9 or fewer people in the sample are left-handed. Before we calculate the z-score, we must find p, the population proportion. We know the total number of left-handed people in the total population, so we know the population proportion: p=4692445=0.19 We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10. We have a sample size of 50 students, so we have: 50(1−0.19)=40.5 40.5 is greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We have μp^=p=0.19; The standard deviation of the sampling distribution of sample proportions isσp^=0.19(1−0.19)50−−−−−−−−−−−−√=0.06 For the given sample of 50 students, the sample proportion isp^=950=0.18. Then, the corresponding z-score is z=p^−μp^σp^=0.18−0.190.06=−0.17 3. Slide the black dot to the z-score we found above: −0.17. The area shown is 0.4325. Rounding to three decimal places, we have P(X≤9)=0.43.
A small community college has a population of 986 students, 252 of whom them are considered non-traditional students. Find the population proportion of non-traditional students, as well as the mean and standard deviation of the sampling distribution for samples of size n=40. Round all answers to 3 decimal places.
Correct answers:1$0.256$0.2562$0.256$0.2563$0.069$0.069 The population proportion is p=252986≈0.256. This is also the mean of the sampling distribution: μp^=p=0.256 For samples of size n=40, the standard deviation of the sampling distribution is σpˆ=p(1−p)n−−−−−−−√=0.256(1−0.256)40−−−−−−−−−−−−−−√≈0.069
A kitchen supply store has a total of 642 unique items available for purchase. Of their available kitchen items, 260 are kitchen tools. The store manager would like to study this further when conducting item inventory. If the store manager surveys 52 store items, what is the probability that 24 or fewer of them are kitchen tools? Round your answers to three decimal places. Use the z-table below:
Correct answers:1$0.405$0.4052$0.462$0.4623$0.068$0.0684$0.838$0.8385$0.800$0.800 We know the total number of items in the total population, so we know the population proportion: p=260642=0.405 The standard deviation of the sampling distribution of sample proportions is thus σp^=0.405(1−0.405)52−−−−−−−−−−−−−−√=0.068 For the given sample of 52 items, the sample proportion is p^=2452=0.462 . The corresponding z-score is z=p^−μp^σp^=0.462−0.4050.068=0.838 Since we are looking for the probability that the number of kitchen tools is less than or equal to 24, we want to know the area under the left tail of the standard normal distribution, to the left of z=0.838. Using the z-table, z=0.84 corresponds to a left-tail probability of 0.800, so P(z≤0.84)=0.800. That is, the probability that 24 or fewer out of 52 items at this store will be kitchen tools is 0.800.
A financial services company is looking to study the types of investments participants make within their online platform. It is known that 56% of the population allocates a quarter of their investments in real estate. If the financial service company randomly samples 45 participants, use the graph below to calculate the probability that more than half of them have a quarter of their investments in real estate. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Enter your answer in decimal form and round to two decimal places.
Correct answers:1$0.79\pm0.001$0.79±0.001 1. We will select the greater than graph option, with the purple area in the right tail of the curve, because we want to know the probability that more than half of the sample size has a quarter of their investments in real estate. That is, for samples of size n=45, we want to find P(p^>0.50), the probability that the sample proportion p^ will be more than 50%=0.50. We are given a population proportion of p=0.56. We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10 and np≥10. We have a sample size of 45 people and p=0.56, so we have: 45(1−0.56)=19.8 and 45(0.56)=25.2. 19.8 and 25.2 are greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We are given that μp^=p=0.56; The standard error is:σp^=0.56(1−0.56)45−−−−−−−−−−−−√=0.0740 We want to calculate the probability of p^=50%=0.50. Then the z-score corresponding to p^=50%=0.50 is z=0.50−0.560.0740=−0.81 3. Slide the black dot to the z-score we found above: −0.81. The area shown is 0.791. Rounding to two decimal places, we have P(p^>0.50)=0.79. So, the probability that more than half of a random sample will have a quarter of their investments in real estate is 0.79, or 79%.
In a video game store, 50% of all visitors in the past 6 months listed multiplayer online games as their preferred type of gaming. If a random sample of 56 visitors are chosen, use the graph below to determine the probability that more than 23 of them prefer multiplayer online games. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find p^ and σp^; Calculate the z-score for p^ and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to two decimal places.
Correct answers:1$0.91\pm0.01$0.91±0.01 1. We will select the greater than graph option, with the purple area in the left tail of the curve, because we want to know the probability that more than 23 people in the sample prefer multiplayer online games. We are given a population proportion of p=0.50. We should also check that we have met the condition for the Central Limit Theorem by plugging in p and n into n(1−p)≥10 and np≥10. We have a sample size of 56 visitors, so we have: 56(1−0.50)=28 and 56(0.5)=28. 28 is greater than 10, so we can use the Central Limit Theorem. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. 2. In order to calculate the z-score, so we need to identify μp^, and calculate p^ and σp^: We know that μp^=p=0.50; We find the standard error to beσp^=0.5(1−0.5)56−−−−−−−−−−√=0.067 We find p^=2356=0.411.So, for samples of size n=56, we want to find P(p^>0.411), the probability that the sample proportion p^ will be more than 41.1%=0.411. Then the z-score corresponding to p^=0.411 is z=0.411−0.500.067=−1.33 3. Slide the black dot to the z-score we found above: −1.33. The area shown is 0.9082. Rounding to two decimal places, we have P(p^>0.50)=0.91. So, to two decimal places, the probability that more than half of a random sample will prefer multiplayer online games is 0.91, or 91%.
Key Terms
Population Proportion: Population Parameter referring to the proportion of subjects in the population that favor a certain characteristic. Sample Proportion: Sample Statistic referring to the proportion of subjects in the sample that favor a certain characteristic Success: The characteristic of interest Trials: Number of people in the sample or population Point Estimate: Sample proportion p^=rn Standard Error of the Sampling Distribution: Standard Deviation σp^=p(1−p)n−−−−−√
Suppose every student in a large school is randomly assigned to one team: team A, team B, team C, or team D. What the probability that in a random sample of 70 students, the proportion of students belonging to team D is between 23% and 28%?
Solution: In this problem, the population proportion of students belonging to team D is 25% since there are 4 equally likely teams. So, p=0.25. We take a random sample of size n=70, and we want the probability that the sample proportion p^ will be between 0.23 and 0.28. That is to say, we want to compute P(0.23≤p^≤0.28). First, let us make sure that the conditions for the Central Limit Theorem for Proportions are met. We know that p=0.25 and n=70, so n(1−p)=52.5 and np=17.5. Since 52.5 and 17.5 are greater than 10, we can proceed with standardizing our variable and using the standard normal table for probabilities. The CLT for sample proportions tell use that p^ is normally distributed with mean μp^=p and σp^=p(1−p)n−−−−−√. In our case, we have μp^=0.25 and σp^=0.25(1−0.25)70−−−−−−−−−−−−√=0.05175. Since p^ is normal, we can compute the probability by reducing it a standard normal random variable by using z-scores: Z1=0.23−0.250.05175=−0.39 Z2=0.28−0.250.05175=0.58 Using the z-table below: We can see that the P(z≤−0.39)=0.348 and P(z≤0.58)=0.719 So, the probability that the sample proportion will be less than 0.28 but more than 0.23 is 0.3707.
Assume that 70% of the people in the United States own a smartphone. What are the mean and the standard deviation of the sampling distribution of the sample proportion of people who own a smartphone, if the sample size is 100?
Solution: The mean of the sample distribution for p^ is the same as the population proportion, 70%, so μp^=0.7. The standard deviation is σp^=p(1−p)n−−−−−−−√=0.7(1−0.7)100−−−−−−−−−−√≈0.046
Sampling distribution of a sample proportion
Suppose simple random samples of size n are drawn from a population with proportion p. Then the sampling distribution of the sample proportions, p^, has: Mean: μp^=p and Standard deviation: σp^=p(1−p)n−−−−−−−√ The standard deviation of the sampling distribution of the sample proportions, σp^, is often referred to as the standard error.