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.

Caption for the picture.

Caption for the picture.

Min 147.2 Q1 163.8 Median 170.3 Mean 171.1 SD 9.4 Q3 177.8 Max 198.1

(a) What is the point estimate for the average height of active individuals? What about the median? (See the next page for parts (b)-(e).)

The point estimate for mean : 171.1 The point estimate for median : 170.3

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

The point estimate for SD : 9.4 The point estimate for IQR : Q3-Q1 = 177.8-163.8 = 14

IQR <- 177.8-163.8

(c) 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.

The approximate 95% confidence interval is ( 171.1 - 2 * SD , 13.65 + 2* SD ) = (152.3, 189.9)

Both can be considered as unusually tall or short since both values are in between the 2 SD confidence range.

confidence_lower_range <- 171.1 -2 * 9.4
confidence_upper_range <-  171.1 + 2 * 9.4

(d) 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.

It cant be same due to the sampling error.

(e) 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 = ! pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Standard error can be used to quantify the variability of the estimate. = SD/sqrt(no of observations) = 0.41

SE <- 9.4/sqrt(507)
SE
## [1] 0.4174687

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.

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

False, the confidence interval can be applied to the population, not the samples.

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

False.Since it is righyly skewed and number of samples are greater than 100, the confidence internal is valid.

(c) 95% of random samples have a sample mean between $80.31 and $89.11.

False, the confidence interval can be applied to the population, not the samples.

(d) We are 95% confident that the average spending of all American adults is between $80.31 and

$89.11.

True.

(e) 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. 90% is narrorer since it has less number of possibilities compared to 95%.

(f) In order to decrease the margin of error of a 95% confidence interval to a third of what it isnow, we would need to use a sample 3 times larger.

False, it have to be nine times, since the Standard error inversly proportional to the square root of the number of observation or samples.

(g) The margin of error is 4.4.

True

mean<- (80.31 +  89.11)/2
margin <- 89.11 - mean

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.

(a) Are conditions for inference satisfied?

Independence -> since it is random sample, it satisfies the independence conditions. Sample Size -> Since the number of samples is more than 30, it satisfies the sample size Distrubition -> The distribution is slighly left skewed, hence inference should be ideal if the number of samples greater than 100. However this is not strongly skewed, hence we can assume that this hold good here.

(b) Suppose you read online that children first count to 10 successfully when they are 32 monthsold, 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.

Current observation of mean = 30.69 months. But the new observation is 32 months.

H0 -> Null Hypothesis => mu = 30.69 HA -> Alternate Hypthesis => mu < 32

calculate the z value of the sampled mean.

z <- (32 - 30.69)/(4.31/sqrt(36))
z
## [1] 1.823666

calculate the pvalue

pvalue <- 1- pnorm(z)
pvalue
## [1] 0.0341013

use the significance level as 0.10 Since pvalue < alpha(0.10), we can reject the null hypothesis.

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

use the significance level as 0.10 Since pvalue < alpha(0.10), we can reject the null hypothesis.

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

Standard error = standard deviation/sqrt(no of samples) confidence level (29.7, 31.6)

std_err <- 4.31/sqrt(36)
z<- abs(qnorm(.1))
lower <- 30.69 - z*std_err
lower
## [1] 29.76942
upper <-  30.69 + z*std_err
upper
## [1] 31.61058

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

Yes, Both matches. The New sampled mean is outside of the confidence and null hypthesis is rejected.

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.

Exercise 4.24

Exercise 4.24

(a) 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 -> Null Hypothesis => mu = 118.2 HA -> Alternate Hypthesis => mu not equals to 118.2

Since the pvalue is 0 which is less than than the significance value(0.10), we can reject H0.

z <- (118.2 - 100)/(6.5/sqrt(36))
z
## [1] 16.8
pvalue <- 2*(1- pnorm(z))
pvalue
## [1] 0

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

Standard error = standard deviation/sqrt(no of samples) confidence level (116.8, 119.5)

std_err <- 6.5/sqrt(36)
z<- abs(qnorm(.1))
lower <- 118.2 - z*std_err
lower
## [1] 116.8117
upper <-  118.2 + z*std_err
upper
## [1] 119.5883

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

Yes, Both matches. The New sampled mean is outside of the confidence and null hypthesis is rejected.

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.

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

6%

mean = 9000
sd = 1000
#using the Z score formula
z <- (10500-mean)/sd
1- pnorm(z)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs. sampling distribution of 10 samples is approximately normal with standard error = 258.1989
n= 15
SampleD <- sd/(sqrt(n))
SampleD
## [1] 258.1989
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? nearly 0%
1 - pnorm(10500,mean = 9000,sd = SampleD)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
normsample <- seq(mean - (3 * sd), mean + (3 * sd), length=1000)
randomsample<- seq(mean - (3 * SampleD), mean + (3 * SampleD), length=1000)
popDist <- dnorm(normsample,mean,sd)
sampleDist<- dnorm(randomsample,mean,SampleD)

plot(normsample, popDist, type="l",col="blue", ylim=c(0,0.008), xlab = "Weights", ylab = "Frequency")
lines(randomsample, sampleDist, col="red")

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

We can not find the probability accuratly for (a) since the distrubution is skewed. We can find the probability for (c) if we have no of samples is greater than 100 since it is skewed. But with samples 10, it is not possible.