## 
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
## This package is designed to support this course. The text book used 
## is OpenIntro Statistics, 3rd Edition. You can read this by typing 
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## The getLabs() function will return a list of the labs available. 
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## The demo(package='DATA606') will list the demos that are available.

Heights of adults. (7.7, p. 260) Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals. The histogram below shows the sample distribution of heights in centimeters.

  1. What is the point estimate for the average height of active individuals? What about the median?

Average height is 171.1. Median is 170.3

  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

SD = 9.4 IQR = Q3-Q1= 177.8-163.8 = 14

  1. Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.
(180-171.1)/9.4
## [1] 0.9468085
(155-171.1)/9.4
## [1] -1.712766
  1. The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning.
  1. The sample means obtained are point estimates for the mean height of all active individuals, if the sample of individuals is equivalent to a simple random sample. What measure do we use to quantify the variability of such an estimate (Hint: recall that \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

It would be the standard error: SE=σ/√n=9.4/√507≈0.417

Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked off on November 27, 2009 (the day after Thanksgiving), had been marked by somewhat lower self-reported consumer spending than was seen during the comparable period in 2008. To get an estimate of consumer spending, 436 randomly sampled American adults were surveyed. Daily consumer spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Determine whether the following statements are true or false, and explain your reasoning.

  1. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

FALSE. Inference is measured on the population parameer. The point estimate of this sample is always within the confidence interval

  1. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

False. This confidence interval is not valid since the distribution of spending in the sample is right skewed.The data is only slightly skewed and the sample is so small this is invalid

  1. 95% of random samples have a sample mean between $80.31 and $89.11.

False.Samples can have different ranges therefor this is incorrect.

  1. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

True.The population parameter is being estimated by the point estimate and the confidence interval.

  1. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

True.With a 90% confidence interval we do not need such a wide interval to catch the values, so the interval would be narrower

  1. In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.

False.In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.It would need to be 9 times larger

  1. The margin of error is 4.4. True.The margin of error is half the confidence interval : 89.11−80.31/2=4.4.

Gifted children, Part I. Researchers investigating characteristics of gifted children col- lected data from schools in a large city on a random sample of thirty-six children who were identified as gifted children soon after they reached the age of four. The following histogram shows the dis- tribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.

  1. Are conditions for inference satisfied? There is not any strong skew in the population. The sample is random and 36 children of a large city is certainly under 10% of the population. Based on this information the conditions for inference are satisfied

  2. Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average age at which gifted children fist count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

H_0 = 32 months H_a = < 32 months Sig level = 0.10

  1. Interpret the p-value in context of the hypothesis test and the data.

z = (30.69-32)/4.31 pnorm(-.3) = 0.38 p_val = 0.38

we fail to reject H_0 per the p_val of 0.38 > 0.1

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

The 90% confidence interval is 30.69±1.65∗SE=30.69±1.188 or (29.502,31.878).

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, the CI of 90% is under 32 months, in what was our hypothesis from the beginning

Gifted children, Part II. Exercise above describes a study on gifted children. In this study, along with variables on the children, the researchers also collected data on the mother’s and father’s IQ of the 36 randomly sampled gifted children. The histogram below shows the distribution of mother’s IQ. Also provided are some sample statistics.

  1. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average IQ of mothers of gifted children is diffrent than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

Null hypothesis: The mean of the mother’s IQ does not differ from the population mean Alternative hypothesis: The mean of the mother’s IQ does differ from the population

(118.2 - 100)/(6.5/6) = 16.8

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
mean = 118.2
n = 36
sd = 6.5
se = sd/sqrt(n)
z_score = (mean -100)/se

low = mean - 1.645*se
up = mean +1.645*se
low
## [1] 116.4179
up
## [1] 119.9821
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Results from the hypothesis test and the confidence interval seem to agree. We are 90% confident that the average IQ of mothers of gifted children is between 116.4 and 120. This is significantly above population average of 100

CLT. Define the term “sampling distribution” of the mean, and describe how the shape, center, and spread of the sampling distribution of the mean change as sample size increases.

The sampling distribution of the mean is the distribution of the mean of samples from the population. If we take random samples from the population and calculate their mean, that distribution will get more normal, the more samples are taken. As sample size increases the normal approximation becomes better and the spread of the sampling distribution of the mean becomes narrower.

CFLBs. A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours.

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
mean = 9000
sd = 1000
prob= (10500 - 9000)/sd
p = 1-pnorm(prob)
p
## [1] 0.0668072
normalPlot(bounds = c(prob,10000000))

  1. Describe the distribution of the mean lifespan of 15 light bulbs.
bulbs = 1000/sqrt(15)
bulbs
## [1] 258.1989

It would be normal

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
score = (10500 - 9000)/258.2
prob = 1 - pnorm(score)
prob
## [1] 3.13392e-09

The probability is 0%

  1. Sketch the two distributions (population and sampling) on the same scale.
norm1 <- rnorm(100000, mean = 9000, sd = 1000)

norm2 <- rnorm (100000, mean = 9000, sd = 1000/sqrt(15))

norms <- data.frame(sample = norm1, means = norm2)

ggplot(norms) + geom_density(aes(x = sample)) + geom_density(aes(x = means))

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

We can not estimate it for part A, but possibly part C depending on the skew. Given survival time for non living things is usually Weibull then yes.

Same observation, different sample size. Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain.

Since we’re increasing sample size to increase the spread of the distribution would be much more narrower, SD would also decrease with a larger n.If n is increased from 50 to 500, then SE will decrease and alternatively Z will increase in case of positive Z and decrease in case of negative Z. As Z is changed, p−value will decreased