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.

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

Answer

heights <- bdims$hgt
paste("Point estimate for the average height of active individuals is ",round(mean(heights),digits=2))
## [1] "Point estimate for the average height of active individuals is  171.14"
paste("Median is",round(median(heights),digits=2))
## [1] "Median is 170.3"

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

Answer

paste("Standard deviation is",round(sd(heights),digits=2))
## [1] "Standard deviation is 9.41"
paste("IQR is",round(IQR(heights),digits=2))
## [1] "IQR is 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.

Answer

Person with 180cm height is around 1 sd, so not considered unusual. Person with height 155cm height is more than 1 sd, so its considered unusual.

(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.

Answer

If the random sampling strategy is similar in both case then there is chance we get similar mean and SD, or else the result can be different.

(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 \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Answer

n <- length(heights)
SD_x = sd(heights)/sqrt(n)
paste("SD_x is", round(SD_x,digits=2))
## [1] "SD_x is 0.42"

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.

Answer

False, Confidence Interval is for the whole population and not just for the 436 random samples

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

Answer

False, Right skew is not prominent here as the significant population is within CI range

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

Answer

False, there is no correlation between confidence interval and sample mean

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

Answer

True, this is based on the confidence interval

(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.

Answer

True, the range gets narrower as the confidence interval becomes smaller.

(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.

Answer

False, to decrease error margin by three times we would need to take sample which is 9 times larger

(g) The margin of error is 4.4.

Answer

True

mrg <- (89.11-80.31) / 2
mrg
## [1] 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.

(a) Are conditions for inference satisfied?

Answer

Conditions of inference are satisfied, the children are selected and the sampling size is > 30. Also the distribution isn’t fully skewed.

(b) 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.

Answer

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

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

Answer

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

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

Answer

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

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

Answer

Yes, the CI of 90% is under 32 months, this 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.

(a) 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.

Answer

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

#summary(gifted$motheriq)
x<-100
n<- 36
sd<- sd(gifted$motheriq)
mean <- mean(gifted$motheriq)
StdErr <- sd/sqrt(n)
Z <- (mean - x)/(StdErr)
paste("Since" , Z,"<>100, we reject the null hypothesis H0")
## [1] "Since 16.7564876528615 <>100, we reject the null hypothesis H0"

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

Answer

paste("Lower is",mean-1.645*StdErr)
## [1] "Lower is 116.38322807943"
paste("Upper is",mean+1.645*StdErr)
## [1] "Upper is 119.950105253903"

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

Answer

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.

Answer

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.

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

Answer

paste("Probability of randomly chosen bulb to last more than 10,500 hrs is", round(1-pnorm(q=10500, mean=9000, sd=1000), digits = 4)*100 ,"%")
## [1] "Probability of randomly chosen bulb to last more than 10,500 hrs is 6.68 %"

(b) Describe the distribution of the mean lifespan of 15 light bulbs.

The lifespan of the light bulbs are nearly normally distributed.

1000/sqrt(15)
## [1] 258.1989

(c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

Answer

score <- (10500 - 9000)/258.2
prob <- 1 - pnorm(score)
paste("Probability is ",prob,", almost 0")
## [1] "Probability is  3.1339197903435e-09 , almost 0"

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

Answer

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), col="red") + geom_density(aes(x = means))

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

Answer

No, without increasing the sample size a skewed distribution will not let us calculate probabilities accurately

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.