DATA 606 Chapter 4 Assignment

4.4 Heights of adults. 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.38 Height 150 160 170 180 190 200 0 20 40 60 80 100 Min 147.2 Q1 163.8 Median 170.3 Mean 171.1 SD 9.4 Q3 177.8 Max 198.1

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

#according to the figures and discription of the data
adult_mean <- 171.1
adult_median <- 170.3
paste("average height is: ", adult_mean)

## [1] "average height is:  171.1"

paste("median height is: ", adult_median)

## [1] "median height is:  170.3"

(b)What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

adult_sd <- 9.4
Q3 <- 177.8
Q1 <- 163.8
adult_IQR <- Q3 - Q1
paste("standard deviation is", adult_sd)

## [1] "standard deviation is 9.4"

paste("IRQ is", adult_IQR)

## [1] "IRQ is 14"

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.

#z value more or less than 2 sd would be considered as tall or short, respectively
adult_sd <- 9.4
adult_mean <- 171.1
z <- (180 - adult_mean)/2*adult_sd
    if (z > 1) {
      print("180 is considered tall")
    } else {
      print("180 is not considered tall")
    }

## [1] "180 is considered tall"

adult_sd <- 9.4
adult_mean <- 171.1
z <-  (adult_mean -155)/(2*adult_sd)
    if (z > 1) {
      print("155 is considered short")
    } else {
      print("155 is not considered short")
    }

## [1] "155 is not considered short"

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. #answer(d): No, I would not expect the mean and the standard deviation between the two groups to be the same. Because the individual is randomly picked, the individual observations would not be the same. But it would expect them to be similar.
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 SDx? = p! n )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

SE <- adult_sd/sqrt(507)
paste("Standard error is: ", round(SE,3))

## [1] "Standard error is:  0.417"

4.14 Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked o??? 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.

We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11. False, we cannot make inference on the population parameter, not on the point estimate. The point estimate is always in the confidence interval.
This confidence interval is not valid since the distribution of spending in the sample is right skewed. False, If the sample size was lower than 30, this would be a problem; however our sample size is 436.
95% of random samples have a sample mean between $80.31 and $89.11. False, the confidence interval is not about a sample mean.
We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. True, this is the definition of a 95% Confidence Interval.
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
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, we would need to make our sample 9 times larger in order to have a confidence interval that is a third of what it is now since we would need the square root of our sample size to be 3 times it’s current value in order to triple the standard error.
The margin of error is 4.4. True, the margin of error is 4.4 since our confidence interval is equal to our point estimate plus and minus the margin of error. (89.11 - 84.71) = 4.4

4.24 Gifted children, Part I. Researchers investigating characteristics of gifted children collected 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 distribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.43

Are conditions for inference satisfied? Yes, (1) samples are random selected. (2) observation is independent. (3) less than 10% of population. (4) size more than 30. (5)no obvious skew on histogram.
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. H0: The average age that gifted children first count to 10 is the same as that of the general population, (=32 months). HA: The average age that gifted children first count to 10 is less than that of the general population, (<32 months).

z_score= (30.69 - 32) / 4.31
z_score

## [1] -0.3039443

library(DATA606)

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normalPlot(mean = 0, sd = 1, bounds = c(-Inf,z_score), tails = FALSE)

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

Because our P value equals to .381 which is greater than our significance level .1, we fail to reject the null hypothesis. There is not enough statistical evidence to support our alternative hypothesis.

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

low_tail <- round(30.69 - 1.65*4.31, 2)
up_tail <- round(30.69 + 1.65*4.31, 2)
low_tail

## [1] 23.58

up_tail

## [1] 37.8

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

4.26 Gifted children, Part II. Exercise 4.24 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.

Perform a hypothesis test to evaluate if these data provide convincing evidence that the average IQ of mothers of gifted children is di↵erent than the average IQ for the population at large, which is 100. Use a significance level of 0.10. H0: The average IQ of mothers of gifted children is the same as that of the general population, (=100)

HA: The average average IQ of mothers of gifted children is different than that of the general population, (!=100)

z_score = (118.2 - 100) / 6.5
z_score

## [1] 2.8

normalPlot(mean = 0, sd = 1, bounds = c(-z,z_score), tails = TRUE)

P = 1-.995
P

## [1] 0.005

P value = .005 is less than our significance level of .1, so we reject the null hypothesis in favor of the alternative hypothesis. There is statistical evidence that the IQ of the mothers of gifted children is different than that of the general population.

Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

low_tail <- round(118.2 - 1.65*6.5, 2)
up_tail <- round(118.2 + 1.65*6.5, 2)
low_tail

## [1] 107.48

up_tail

## [1] 128.93

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

Yes, the average IQ of mother’s in the general population (100) does not fall within the 90% confidence interval of the mean IQ of mothers of gifted children (107.48 to 128.93).

4.34 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.

Sampling distribution of the mean is the distribution of the mean from many different samples from a population. As the sample size increases the sampling distribution becomes closer to the Normal Distribution in its shape. The center does not change much and approaches the population mean. The spread narrows as sample size increases.

4.40 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.

What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

z_score <- (10500-9000)/1000

normalPlot(mean = 0, sd = 1, bounds = c(-Inf,z_score), tails = FALSE)

p <- 1-0.933
p

## [1] 0.067

So the probability that a randomly chosen light bulb lasts more than 10,500 hours is 0.067

Describe the distribution of the mean lifespan of 15 light bulbs. Could be anything because the sample size is not big enough to predict any outcome; central limit theorom applies, if anything it should be near normal distribution.
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

1-pnorm(10500,mean=9000,sd=258)

## [1] 3.050719e-09

Sketch the two distributions (population and sampling) on the same scale.

x = 7000:11000
y = dnorm(x,mean=9000,sd=1000)
X = 7000:11000
Y = dnorm(X,mean = 9000,sd=258)
plot(x,y)

plot(X,Y)

Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? if the sample size is big enough, it is possible to estimate the probabilities. If the sample size is too small, it will be hard.

4.48 Same observation, di↵erent 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. statand error gets smaller as sample size gets bigger. p-value will descrease if you increase the sample size.

DATA 606 Chapter 4 Assignment

Jun Pan

October 14, 2018