Normal and Sampling Distributions - part 2

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^p denote the proportion in the sample who say they support the increase. Suppose that 79% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

= SQRT(0.79*(1-0.79)/1500) = 0.0105

sampling distribution for sample mean

= sd/SQRT(n)

Central Limit Theorem

The theory that, as sample size increases, the distribution of sample means of size n, randomly selected, approaches a normal distribution. - (just rule of thumb: always assume x, or the sample is approximately normal when it is greater than 30)

mean for sample proportion

ˆp has a normal distribution with a mean of μˆp = p

standard deviation for sample proportion

σˆp= SQRT(prob*(1-prob /n) (and as long as np and n(1 - p) are at least 10).

A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 4.20 mm?

x = 4.2 mean = 4 sd = 1.1 n = 100 - find the sample mean first = sd/sqrt(n) = 0.11/sqrt(100) = 0.11 = norm.dist(4.2, 4, 0.11, 1) = 0.96548

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .2005 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean

- mean = 16.05 - sd = .2005 - n = 4 - find the sample standard deviation = .2005/sqrt(4) =.10025 apply stdrd deviation rule - 95% = 2SD = 16.05 - 2*0.10025 = 15.8495 = 16.05 + 2*0.10025 = 16.2505

Determine the parameter and statistic in the scenario, and explain: The SAT-Verbal scores of a sample of 300 students at a particular university had a mean of 592 and a standard deviation of 73. According to the university's reports, the SAT-Verbal scores of all its students had a mean of 580 and a standard deviation of 110.

- parameter: standard deviation of 110, the scenario states that these are the scores for ALL university students - which would make it a population standard deviation and describes the population - statistic: mean of 592, the mean is what we got from a sample of 300 students

Suppose that 78% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places?

- we would use norm.dist for this problem but we have to find the standard deviation first =SQRT(.78*(1-.79)/100) = 0.041425 (don't round) = 1 - norm.dist(.83,.80, 0.037563, 1) = 0.685383

Suppose we take repeated random samples of size 20 from a population with a mean of 60 and a standard deviation of 8. Which of the following statements is true about the sampling distribution of the sample mean (x̄)? Choose all that apply. a.) The distribution is normal regardless of the shape of the population distribution, because the sample size is large enough. b.) The distribution will be normal as long as the population distribution is normal. c.) The distribution's mean is the same as the population mean 60. d.) The sampling distribution's standard deviation is larger than the population standard deviation of 8.

Ans: a & c, as long as the sample is greater than 30 we can assume the distribution is normal, and the mean will be the same as the population mean.

sample distribution

a frequency distribution of a sample

Statistic

a number that describes the sample, basically the numbers you get from the sample

In June 2015, Gallup conducted a poll of a random sample of 14761 adults to determine the well-being of people living in the United States. One question asked, "Did you exercise at least 30 minutes for 3 or more days in the past week?" In the survey, 55.3% of males and 44.7% of females responded yes to this question. Which of the following is true about this scenario? a.) 55.3% and 44.7% are both statistics. b.) If we took another random sample of 14761 adults, we would expect to get the exact same results. c.) 55.3% and 44.7% are both parameters.

a.) both are statistics because it's a numerical description of sample

Which of the following statements about the sampling distribution of the sample mean, x-bar, is true? Check all that apply. a. The distribution is normal regardless of the shape of the population distribution, as long as the sample size, n, is large enough. b. The distribution is normal regardless of the sample size, as long as the population distribution is normal. c. The distribution's mean is the same as the population mean. d. The distribution's standard deviation is smaller than the population standard deviation

all are correct!

Suppose that a candy company makes a candy bar whose weight is supposed to be 50 grams, but in fact, the weight varies from bar to bar according to a normal distribution with mean μ = 50 grams and standard deviation σ = 2 grams. If the company sells the candy bars in packs of 4 bars, what can we say about the likelihood that the average weight of the bars in a randomly selected pack is 4 or more grams lighter than advertised? a.)There is about a 16% chance of this occurring. b.) There is about a 5% chance of this occurring. c.) There is about a 2.5% chance of this occurring. d.) There is no way to evaluate this likelihood, since the sample size (n = 4) is too small. e.) It is extremely unlikely for this to occur; the probability is very close to 0.

ans: e - mean = 50 - sd = 2 - x = 4 P(x<46) = (because the problem is asking for a pack of four grams lighter than the average weight of 50 advertised) - find the sample standard deviation = 2/sqrt(4) = 1 = norm.dist(46, 50, 1, 1) = the correct answer is 3.1617 E - that's scientific notation for zero!

The number of hours a light bulb burns before failing varies from bulb to bulb. The distribution of burnout times is strongly skewed to the right. The Central Limit Theorem says that a.) the average burnout time of any number of bulbs has a distribution that is close to Normal. b.) the average burnout time of a large number of bulbs has a distribution of the same shape (strongly skewed) as the distribution for individual bulbs. c.) the average burnout time of any number of bulbs has a distribution of the same shape (strongly skewed) as the distribution for individual bulbs. d.) as we look at more and more bulbs, their average burnout time gets ever closer to the mean for all bulbs of this type. e.) the average burnout time of a large number of bulbs has a distribution that is close to Normal.

ans: e, satisifies the central limit theorem - assuming a large distribution is normal

Suppose the American National Elections Studies agency (ANES) wishes to conduct a survey. It plans to ask a yes/no question to determine if those surveyed plan to vote for a certain candidate. One proposal is to randomly select 400 people and another proposal is to randomly select 1600 people. Which of the following is true regarding the sample proportion ^p of "yes" responses? a. The sample proportion from the sample of 400 is more likely to be close to the true population proportion, p. b. The sample proportion from sample of 1,600 is more likely to be close to the true population proportion, p. c. The sample proportion, ^p, in either proposal is equally likely to be close to the true population proportion, p, since the sampling is random.

b.) a sample proportion of 1600 has a bigger sample size, it'll be easier to gauge a true population proportion by having

Parameter

describes the population, for example: population proportion, sample mean, population standard deviation - all these describes the population or a sample from the population

Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a simple random sample (SRS) of 5 students from this distribution. What is the probability that a SRS of 5 students will spend an average of between 600 and 700 dollars? Round to five decimal places.

sample mean = 650 sd = 120 n = 5 to find the sample mean between 600 and 700 students, find the sample mean manually for both value - 120/sqrt(5) = 53.66563 x = (600-650)/53.66563 = (-.93169) x = (700-650)/53.66563 = .93169 = norm.s.dist(0.93169, 1) - norm.s.dist(-.93169, 1) = 0.64851

Sampling distribution of the sample proportion

the distribution of sample proportions, with all samples having the same sample size (n) taken from the same population