A random survey was conducted to estimate the proportion of people who favoured reading using an e-reader over a standard book. It was found that 286 out of 419 people surveyed preferred the e-reader.
Find the sample proportion \hat{p} of those in the survey who preferred to use an e-reader. Round your answer to two decimal places.
Working with a two-sided confidence interval of 90\%, estimate the minimum sample size necessary to ensure a margin of error of at most 0.05 if the sample proportion remains the same.
Using the sample proportion \hat{p} from the initial survey, the 95\% confidence interval for p is 0.64 \leq \hat{p} \leq 0.72. Considering this interval, which of the following surveys are more likely to be representative of the total population?
A random sample of 79 at a book store found that 31 had a preference for e-readers.
A random sample of 365 at an inner city park found that 256 had a preference for e-readers.
In a sample of 350 people, it is found that only one has blood type B-negative. Let p represent the proportion of the population that have blood type B-negative.
Find an estimate for p.
Find an approximate two-sided 95\% confidence interval for p. Round your values to four decimal places.
Interpret the confidence interval found in part (b).
One measure of the validity of a confidence interval is that the product of the sample size n and the population proportion p is greater than 5. Estimate this product for the blood type sample.
Explain the meaning of the result in part (d) with regards to the validity of the confidence interval.
Jimmy works on the top floor of a 50-storey building. The probability that the elevator will stop at another floor on its way up to his office is p. Jimmy has decided to test this probability by noting the outcome for every one of the 236 working days of the year, over five years. He records a 1 if the elevator does stop, and a 0 if it doesn't stop.
The average outcome for each year is shown in the given table:
| Year | Sample Average |
|---|---|
| 1 | 0.71 |
| 2 | 0.67 |
| 3 | 0.71 |
| 4 | 0.78 |
| 5 | 0.65 |
Calculate an approximate two-sided 95\% confidence interval for p using the sample average from Year 1. Round your values to three decimal places.
How confident can we be that p is inside this interval"
Calculate an approximate two-sided 95\% confidence interval for p using the sample average from Year 4. Round your values to three decimal places.
30 hamburger patties advertised as being 180 \text{ g} are weighed and the results are tabulated:
What is the sample proportion of patties weighing less than 180 \text{ g}?
Use this sample to calculate an approximate two-sided 95 \% confidence interval for patties weighing less than 180 \text{ g}. Round your values to two decimal places.
A second sample of patties was similarly weighed and a 95 \% confidence was calculated as \left(0.4019, 0.6781\right). How many patties were in this sample? Round your answer to the nearest whole number.
| \text{Hamburger} \\ \text{patty mass (g)} | \text{Number of patties} |
|---|---|
| 178\leq w\lt179 | 7 |
| 179\leq w\lt180 | 8 |
| 180\leq w\lt181 | 8 |
| 181\leq w\lt182 | 5 |
| 182\leq w\lt183 | 2 |
Let p be the proportion of 5 \text{ kg} bags of potatoes that actually weigh less than 5 \text{ kg}. A sample of 400 such bags were weighed and a confidence interval for the sample proportion was calculated as \left(0.26, 0.32\right).
How many bags of potatoes in this sample weighed less than 5 \text{ kg}?
If 60 samples, each containing 400 of the 5 \text{ kg} bags were selected, and the two-sided 95\% confidence interval for each sample was calculated in the same manner, approximately how many of these confidence intervals would contain p?
A random sample of 100 people indicated that 19\% had visited an art gallery in the past year.
Determine an approximate two-sided 95\% confidence interval for the proportion of the population that had visited an art gallery in the past year. Round your values to two decimal places.
A new sample of 100 people was taken. Let x be the number of people in the sample who had visited an art gallery in the past year. Give a range, using a 95\% confidence interval, within which you would expect X to lie. Round your values to the nearest whole number.
Determine the probability that, in a random sample of 120 people, the number who had visited an art gallery in the past year was greater than 22. Round your answer to four decimal places.
Determine the probability that, out of five random samples, at least four 95\% confidence intervals include the true value of p. Round your answer to four decimal places.
A ride-sharing company took a random sample of 270 of their recent customers in the Sydney CBD and found that 20 of them were dissatisfied with their experience using their services.
Calculate a two-sided 99 \% confidence interval for the proportion of Sydney CBD customers who would be dissatisfied with their experience. Round your values to two decimal places.
Similar samples show that 6\% of customers using their service in CBDs all over Australia are dissatisfied with their experience. Using the confidence interval, is there evidence to suggest that the Sydney CBD customers are more dissatisfied than average? Explain your answer.
A survey in 2012 found that 8.7 \% of Victorians are vegetarian. A more recent study in 2016 of 1800 Victorians found that 196 people had a vegetarian diet.
Using the 2016 data, construct a 95 \% confidence interval for the proportion of Victorians eating a vegetarian diet. Round your values to three decimal places.
Is there evidence to support a conclusion that the proportion of people in Victoria that eat a vegetarian diet has increased since 2012? Explain your answer.
A producer of instant scratch lottery tickets claims that 5 \% of its tickets win a prize. A consumer group wants to test this claim. In a randomly selected sample of 500 tickets, 19 were found to be winning tickets.
Construct an approximate two-sided 95 \% confidence interval for proportion p of winning tickets. Round your values to three decimal places.
Is there evidence to refute the manufacturer's claim? Explain you answer.
It was found that 85 \% of Australians do not apply enough sunscreen before outdoor activities. In an attempt to reduce this percentage, a public awareness campaign was launched to inform people on the correct amount to apply and when.
The following summer, a new survey was conducted to assess whether the campaign was effective. 362 people out of 450 surveyed did not apply the correct amount of sunscreen.
Construct an approximate two-sided 95 \% confidence interval for the proportion of the population who did not use the correct amount of sunscreen. Round your values to three decimal places.
Is there evidence to refute a claim that the campaign was ineffective? Explain your answer.
A supermarket is surveying individuals to determine what proportion is moving towards online grocery shopping. A random sample is taken of 205 people, and 80 people regularly shop online.
Determine the sample proportion \hat{p} of those in the survey who preferred to grocery shop online.
Use the survey results to estimate the standard deviation of the sample proportions, correct to four decimal places.
A follow up survey is to be conducted to confirm the results of the initial survey. Working with a confidence interval of 95 \%, estimate the minimum sample size necessary to ensure a margin of error of at most 0.05 for the same sample proportion.
It is known, from the last census, that 21 \% of Australians speak a language other than English at home. The ABS takes 10 random samples and calculates a point estimate and 90 \% confidence for each. The confidence intervals are tabulated:
How many confidence intervals do not contain the true population proportion?
Explain why we expect to see approximately 1 out of the 10 confidence intervals without the true population proportion.
If the ABS take a further 8 samples, what is the probability that all 90 \% confidence intervals will contain the true population proportion?
| Sample | Confidence interval |
|---|---|
| 1 | (0.011,0.256) |
| 2 | (0.111,0.356) |
| 3 | (0.178,0.422) |
| 4 | (0.044,0.289) |
| 5 | (0.011,0.256) |
| 6 | (0.111,0.356) |
| 7 | (0.078,0.322) |
| 8 | (0.144,0.389) |
| 9 | (0.211,0.456) |
| 10 | (0.011,0.256) |
A student performs an experiment in which a computer is used to draw a random sample of size n from the first 100 integers greater than zero. The proportion of this population of integers that are prime numbers is p.
Let the random variable \hat{P} represent the sample proportion observed in the experiment. If p = 0.25, find the smallest integer value of the sample size such that the standard deviation of \hat{P} is less than \dfrac{1}{100}.
Each of 25 students in a class performs the experiment and calculates an approximate 95\% confidence interval for p using the sample proportions for their sample. It is found that exactly one of the 25 confidence intervals calculated by the class does not contain the value of p.
Two of the confidence intervals from the class are selected at random. Find the probability that exactly one of the selected confidence intervals does not contain the value of p.
A coin is tossed 150 times and 84 heads appeared.
State the sample proportion \hat{p} of the number of tosses that turned up heads.
If the coin was fair, what would you expect the population proportion to be?
Construct an approximate two-sided 95 \% confidence interval for the population proportion of tosses showing heads for this coin. Round your values to two decimal places.
Is there evidence to support a claim that the coin is biased? Explain your answer.
A six-sided die is rolled 200 times and 20 threes appeared on the uppermost face.
State the sample proportion \hat{p} of the number of times a three was rolled.
If the coin was fair, what would you expect the population proportion to be?
Construct an approximate two-sided 95 \% confidence interval for the population proportion of rolls showing a three for this die. Round your values to three decimal places.
Is there evidence to support a claim that the die is biased? Explain your answer.
A chocolate company claims that 24 \% of their chocolate drops are blue. Quiana buys a packet to test the claim, and out of 210 candies, 49 were blue.
State the sample proportion \hat{p} of the number of blue chocolate drops.
Construct an approximate two-sided 95 \% confidence interval for the population proportion of blue chocolate drops. Round your values to three decimal places.
Is there evidence to refute the claim? Explain your answer.
Quiana is not convinced and buys a larger bag. This time, out of 2140 chocolate drops, 472 were blue. Does this sample offer evidence to refute the claim at a 95 \% confidence level? Explain your answer.
A farmer bought a large bag of seeds. She planted 100 of the seeds from this bag and observed that 53 of these seeds were viable.
Calculate the sample proportion of planted seeds that were viable.
Construct an approximate two-sided 95 \% confidence interval for the proportion of the seeds that are viable. Round your values to three decimal places.
The distributor of the seeds claims that, on average, 60 \% of seeds are viable. Is there evidence to refute this claim? Explain your answer.
The proportion of the population of the United States thought to have Celiac disease is p. A sample of 2000 Americans were surveyed for the disease and a confidence interval for the sample proportion was calculated as \left(0.0089, 0.0121\right).
How many people in this sample had the disease?
Use the margin of error to find the z-score for this confidence interval, correct to three decimal places.
What is the level of confidence for this sample? Give your answer as a whole percentage.
A tea company are surveying supermarkets to determine their most popular flavour of tea. Sample A has 420 recent sales, 240 of which were English Breakfast.
Calculate a two-sided 90 \% confidence interval for Sample A. Round your values to two decimal places.
In a second sample, B, of 420 sales, the two-sided 99 \% confidence interval for English Breakfast sales was approximately \left(0.4730, 0.5984 \right). Calculate \hat{p} for the second sample.
Which sample has a larger margin of error? Explain your answer.
An online book store takes two random samples of size 50 from their book sales to determine the proportion of customers buying digital copies, rather than buying soft or hard covers. Sample A has confidence interval \left(0.20 , 0.46 \right). Sample B has confidence interval \left(0.11, 0.34 \right), with the same level of confidence as Sample A.
Calculate \hat p for Sample A.
Calculate \hat p for Sample B.
Which sample has a larger margin of error? Explain your answer.
To estimate the true proportion of residents in a particular city who think that a new train line needs to be added to the existing network, samples were taken and the proportion of those who agreed was calculated.
In a sample of 180 residents, the proportion who agreed was 0.76 and the margin of error associated with the two-sided confidence interval calculated was 0.05. Calculate the level of confidence for this sample, to the nearest percent.
A pharmaceutical company wants to ascertain the proportion of the population that carry a particular gene that can heighten the risk of lung cancer.
Sketch the graph of y = x \left(1 - x\right) on a number plane.
Hence, find the value of \hat{p} that maximises the margin of error for a two-sided 95 \% confidence interval.
If the company has no current estimate for the proportion of people with the gene but want to create a two-sided 95 \% confidence interval within a margin of error of 2 \%, how many people will they need to survey to guarantee this?
If a preliminary study suggests 12 \% have the gene, how many would they need to survey to create a two-sided 95 \% confidence interval with no more than a 2 \% margin of error?
A medical facility wants to ascertain the proportion of the population of Australia that have a vitamin D deficiency.
If the facility has no current estimate for the proportion of people with the gene but want to create a two-sided 95 \% confidence interval with a margin of error no more than 4 \%, how many people will they need to survey to guarantee this?
Through previous studies, the Australian Bureau of Statistics estimates that 23 \% of Australians have a vitamin D deficiency. Using this estimate, how many people would the medical facility need to survey to create a two-sided 95 \% confidence interval with no more than a 4 \% margin of error?
A current medication for asthma has a 75 \% success rate in reducing the number and severity of attacks in patients. A drug company claims their new medication has a higher success rate. How many asthma sufferers would need to be included in a study to create a 95 \% confidence interval, with a width of no more than 2 \%, to help assess this claim?
Roxanne and Emma run an ice cream shop and they are wondering whether they should also sell coffee at their shop.
Roxanne thinks that the proportion of customers who would buy coffee is 0.3.
Calculate the minimum size of the sample required for Roxanne to achieve a margin of error of less than 0.06 in an approximate two-sided 90\% confidence interval for this proportion.
Emma thinks that the proportion of customers who would also buy coffee is 0.4.
Calculate the minimum size of the sample required for Emma to achieve a margin of error of less than 0.06 in an approximate two-sided 90\% confidence interval for this proportion.
Suppose that Emma's estimate of the proportion of customers wanting coffee is determined from a sample using the sample size from Roxanne's estimate. What will be the margin of error for the actual proportion p, using a 90\% confidence interval? Round your answer to three decimal places.
A newspaper article makes the following claim:
"Four-fifths of Australian parents say they are too tired after work to do some of the things they would like to do with their children."
A survey is undertaken to investigate the impact of work on family life. Of the 520 parents surveyed, 384 state that they are too tired after work to do some of the things that they would like to do with their children.
Construct an approximate two-sided 95 \% confidence interval for the proportion of parents who are too tired to do some of the things that they would like to do with their children. Round your values to three decimal places.
Is there evidence to refute the article's claim that four-fifths of Australian parents say they are too tired after work to do some of the things that they would like to do with their children? Explain your answer.
What is the minimum number of parents to survey that will result in an approximate two-sided 99 \% confidence interval with width of no greater than 0.06?