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?

Ans. Point Estimate Mean is 171.1 and Median is 170.3.

summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

IQR= 3rd Qrt-1st Quartile

sd(bdims$hgt)
## [1] 9.407205
177.8 - 163.8
## [1] 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.

180CM is within .94 SD away from the mean, which means its not unusually tall. The 155 cm is 1.7 sd away from the mean which is unusually short as its almost over 90th percintile.

(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.

The mean and SD should be the similar as the sample size is large enough.

  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.

We would usestandard error which is the measure of variability of all variables estimates. SE= SD/Sqrt(n)

9.4/sqrt(507)
## [1] 0.4174687

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- Sample mean always fall within CI.

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

True- The distribution is slightly skewed so the CI should not be valid.

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

True, Sample mean should be with in CI of the orignal sample.

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

True, The sample is large enough to make this assumption.

  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, the larger the sample or the lower the CI then the narower the CI.

  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. We need a sample size 9 times bigger.

  1. The margin of error is 4.4.

True. (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?

Yes, Sample is large enough and random. The population distribution is not strongly skewed so is normal. and the sample are independent.

  1. 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.
n=36; x=30.69; sd=4.31
se=sd/sqrt(n)
zscore=(x-32)/se
p=pnorm((zscore))
p
## [1] 0.0341013
  1. Interpret the p-value in context of the hypothesis test and the data.

We reject the null hypothesis as p< .10

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
z=1.645
a=x-z*se
b=x+z*se
a
## [1] 29.50834
b
## [1] 31.87166
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, our results agreee as tge 32 months is outside our CI.

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. Performahypothesistesttoevaluateifthesedataprovideconvincingevidencethattheaverage 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..

H0: Average IQ of mothers of Gifted children = 100 H1: Average IQ of mother of Gifted children != 100

P value of 0< .1, we reject the null hypotheses. There is Significant evidence that the avergae IQ of mothers of gifted chikdreb is different fom the rest of population.

x=118.2; sd=6.5; n=36
zscore=(x-100)/sd*sqrt(n)
se=sd/sqrt(n)
se
## [1] 1.083333
zscore
## [1] 16.8
1-pnorm((zscore),2)
## [1] 0
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
x-1.64*se
## [1] 116.4233
x+1.64*se
## [1] 119.9767
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes agree, averageIQ if 100 is not in the CI of 116.4 and 119.98

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.

Samling distribution means that we aare taking a sample from a distribution which maybe or maybe not normal. However as the size od the sample that we take from the distribution increase the samle mean approaches the true population mea and the shape of it become a normal distribution with the spread becoming leaner and leaner as sample size increase.

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?

Probablity = 6.68%

x=9000
sd=1000
z=(10500-x)/sd
p=1-pnorm(z)
p
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.

Distribution is normal with N(9000,258.2)

se=sd/sqrt(15)
se
## [1] 258.1989
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

0% as the z score is 5.80 away from the mean. There in no chance that 15 randomly choses lights will last more than 10500.

z= (10500-x)/se 
z
## [1] 5.809475
  1. Sketch the two distributions (population and sampling) on the same scale.
x <- seq(5000, 13000,length=sd)
plot(x,dnorm(x,mean=9000,sd=sd),type = "l",lty=1,lwd=3,col="blue", ylim = c(0,.0015))
x <- seq(8000,10000,length=200)
curve(dnorm(x,9000, 258.2),add=TRUE,lty=2,col="red")

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

No we couldnot estimate with a skewed distribution becasue sample size is not large enough. Normallt for CLT, we would like minumum n>30.

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.

The z score will decrease and the p value will decrease as the sample szie increase thus you might have a different result.