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 8 Q1 13 Median 14 Mean 13.65 SD 1.91 Q3 15 Max 18

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

The average height is 171.1. The median is 170.3

  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

The SD is 9.4. IQR = 177.8 - 163.8 = 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 1 SD of the mean. It is taller than the average, but not unusually tall. On the other hand, 155cm is almost 2 SD below the mean, so it is relatively short, but is still falls within the 97.5% range.

mu <-171.1
one.sd <- 9.4
# 1 SD above the mean
mu-one.sd
## [1] 161.7
# 2 SB below the mean
mu-2*one.sd
## [1] 152.3
  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.

No. Point estimates for samples will be different due to sampling variation

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

Using the \({ SD }_{\overline { x }}=\frac {\sigma}{ \sqrt{ n } }\) formula and n=507, we get the following value for SE 

SE <- one.sd/sqrt(507)
round(SE,2)
## [1] 0.42

4.14

Thanksgiving spending 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.

This statement is false because we are making an inference on a point estimate

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

False: While the graph is skewed to the right, the majority of the values appear nearly normal. We also have a large sample of 436

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

False. confidence intervals state that we are 95% confident that we capture the true mean

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

This statement is true since we are referencing the average

  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.

This is true. a 90% confidence interval has a smaller z score

  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. To decrease the margin of error, we would need to sample 9 times what it currently is

  1. The margin of error is 4.4.

True. 4.4 is equal to z*SE

n <- 436
z.times.se <- 4.4
z<-1.96
round(z.times.se/z,2)
## [1] 2.24

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.

#given
n <- 36
min <- 21
mean <- 30.69
sd <-4.31
max <-39
  1. Are conditions for inference satisfied?

We need to check the following conditions: 1. Random, 2. Normal, 3. Independent. Based on the sample size 36, the independent condition is met due to \(n\ge 30\). Based on the give graph, the population is not strongly skewed. We do not know what the total population is, but asssuming that 36 is less than 10%, the condition should be satisfied

  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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

We assuming that the gifted children’s numbers are not any different from the norm. H0 = The gifted children’s reading is the same as the average. H1 = the gifted children’s reading is lower than the average.

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

Using 32 as x. We need to calculate the SE. This is SD/sqrt(N) = 0.72. Z is approx -1.82. Using the pnorm function, we can see that this value is 0.0344. Since this value is less than the significance level 0.10, we can reject the null hypothesis. 

se <- round(4.31/sqrt(36),2)
x.bar <- 32
x.observed <- 30.69
z <- ((x.observed - x.bar)/se)
round(z, 4)
## [1] -1.8194
round(pnorm(z),4)
## [1] 0.0344
  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully. For this interval, we will use the z value for a one tail test. z= 1.28. If we were using a 2 tail test, z=1.65 \(\overline { x } ={ z }^{ * }{ SE }_{ \overline { x } }\). We get 30.69+-1.28*se. We see that our confidence interval is (31.08,32.92).
# Using a one-tail test .90 confidence interval
upper <- 32+1.28*se
lower <- 32-1.28*se
round(upper,2)
## [1] 32.92
round(lower,2)
## [1] 31.08
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The observed mean is still lower than the lower interval, so we can still reject the 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.

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

H0: Mother’s IQ is the same as the standard population H1: Mother’s IQ is different as the standard poluation

se <- round(6.5/sqrt(36),2)
x.bar <- 100
x.observed <- 118.2
z <- ((x.observed - x.bar)/se) 
round(z, 4)
## [1] 16.8519
round(2*pnorm(z, lower.tail = FALSE),8)
## [1] 0

We get a pvalue of 0.0. This means that we can reject the null hypothesis because there is basically a 0 percent chance that we would observe these values during normal sampling

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
# Using a tail-tail test .90 confidence interval, our z score is 1.64
upper <- x.observed+1.64*se
lower <- x.observed-1.64*se
paste0("Upper Range: ", round(upper,2))
## [1] "Upper Range: 119.97"
paste0("Lower Range: ",round(lower,2))
## [1] "Lower Range: 116.43"
paste0("Confidence Interval (", round(lower,2), ", ",round(upper,2), ")")
## [1] "Confidence Interval (116.43, 119.97)"
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The results do agree as the confidence interval for the mothers of the gifted childred is (116.43, 119.97). The general popluation observed mean is 100. We have are 90% certain that we have captured the true mean of the gifted mothers from 116 to rought 119

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 is based on random samples from a sample population. The shape is nearly normal as the sample size increases and the spread decreases as the sample size increases. The larger the sample size, the closer we converge towards the true mean

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? We need to find the z value of 10500. Based on the z value, the probability is roughly 6.7%
mu <- 9000
sd <- 1000
p.observed <- 10500
z <- (p.observed-mu)/sd
z
## [1] 1.5
round(pnorm(z, lower.tail = FALSE),3)
## [1] 0.067
  1. Describe the distribution of the mean lifespan of 15 light bulbs. We are given the SD for the population: 1000. Therefore, we can compute the SE for a sample of 15 using the following formula $SE= $
SE <- 1000/sqrt(15)
round(SE,2)
## [1] 258.2
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

The probability is pretty much zero

  1. Sketch the two distributions (population and sampling) on the same scale.
xvals <- seq(5000,15000, by=100)
fx.pop <- dnorm(xvals, mean = mu, sd=sd)
plot(xvals, fx.pop, type="l", xlab="",ylab = "", ylim=c(0,.0015))

xsample <- seq(5000,15000, by=100)
fx.sample <- dnorm(xvals, mean = mu , sd=SE)
lines(xvals, fx.sample, type="l", xlab="",ylab = "",lwd=2)

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? With a skewed graph, i don’t believe it would be easy to estimate the probability without a nearly normal distribution</blue ### 4.48

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. With a larger sample size, our SE decreases. Using problem 4.27 as an example, we will take the given values 1.75, sample size 110. The SE is .17. The observed mean is 7 and the sample mean is 7.42, which gives us a Z value of 2.47. If we decrease the SE, our Z value will become larger, which means that the probability will become smaller when substracted from 1