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.

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).

Point estimate for average height of active individuals - 171.1.Point estimate for median height -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 height of active individuals = 9.4.IQR = (Q3-Q1)= 14

(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.

x <- 180
mu <- 171.1
sd <- 9.4
z <- (x-mu)/sd
z1 <- (180-171.1)/(9.4)
z1
## [1] 0.9468085
z2 <-  (155 - 171.1)/(9.4)
z2
## [1] -1.712766
As z<2,the consideration is not unusally high or unusually short.

(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.

NO I would definately not expect the sample have same values as above.

(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 weuse to quantify the variability of such an estimate (Hint: recall that SDx ̄ = pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

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

4.14) 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.

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

FALSE,Reason : From the example,the 100% confidence interval ranges from(80.31 - 89.11)

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

Reason: False, n> 30 and here n= 436 ,the condition is satisfied.

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

False,100% sample is between above said range.

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

True,confidence interval is 95%.

(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,since we dont need to be sure about our estimates,we can lower confidence interval.

(f) 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,reducing margin of error three times,the sample size is reduced 9 times.

(g) The margin of error is 4.4.

True.
#Margin_error <- (uppertail - lowertail)/2
Margin_error <- (89.11 - 80.31)/2
Margin_error
## [1] 4.4

4.24) 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.

n 36 min 21 mean 30.69 sd 4.31 max 39

(a) Are conditions for inference satisfied?

Conditions checked 1) n >= 30,here n = 36,hence check.
                    2) Data is independent
                    3) Does not appear to be strongly skewed.Hence conditions are satisfied.

(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.

a) null hypothesis: the average development for child is mu = 32.
b) Alternate hypothesis : average development for child is not 32 months.now lets find p-value.
SE <- 4.31/sqrt(36)
SE
## [1] 0.7183333
z <- (30.69 - 32)/(SE)
p <- pnorm(z,mean = 0,sd = 1)
p* 2
## [1] 0.0682026

Hence we reject null hypothesis.(H0)

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

significance level <- 0.10.P value we got is lower than significance level,we can conclude that gifted children count 10 times faster than normal child

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

z value for 90% confidence interval is 1.65
UpperTail <- round(30.69 + 1.65 *(0.718),2)
LowerTail <- round(30.69 - 1.65 *(0.718),2)
UpperTail
## [1] 31.87
LowerTail
## [1] 29.51

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

Yes,the result agree as we reject null hypothesis.

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.

n 36 min 101 mean 118.2 sd 6.5 max 131

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

here x <- 100,significance level <- 0.10 Null Hypothesis(H0) : The average of mother’s IQ of gifted children = population’s IQ average

Alternate Hypothesis(HA) : The average of mother’s IQ of gifted children != Population IQ average

If p-value != significance level ,we reject null hypothesis. to find p-value

SE <- 6.5/sqrt(36)
z <- (118.2 - 100)/(SE)
p <- (1-pnorm(z,mean = 0,sd = 1))
p
## [1] 0

as p-value = 0,we accept Alternate hypothesis.

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

UpperTail <- round(118.2 + 1.65 *(SE),2)
LowerTail <- round(118.2 - 1.65 *(SE),2)
UpperTail
## [1] 119.99
LowerTail
## [1] 116.41

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

Yes,as we reject null hypothesis,the results do agree.

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.

The sampling distribution of mean is distribution of mean values from samples of population.narrower the spread is,more are the samples involved.shape is more close to normal if more is the number of samples taken which also shifts center.

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.

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

p<- round(1-pnorm(10500,mean = 9000,sd = 1000),4)
p
## [1] 0.0668