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?

Here we interpret the true population as all physically active individuals, and the sample (from which the estimates come from) as the bdims dataset

#Point esitmate for mean height among active individuals:
meanhgt<- mean(bdims$hgt)
meanhgt
## [1] 171.1438
#Median:
median(bdims$hgt)
## [1] 170.3
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
#Standard Deviation:
sdhgt <- sd(bdims$hgt)
sdhgt
## [1] 9.407205
#IQR
IQR(bdims$hgt)
## [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.

‘Unusual’ is pretty subjective, but assuming it means greater than 1 stdev away from the mean: A 180cm tall person would not likely be considered unusually tall, as they are within one stdev (9.4) of the mean height (171.1). A 155cm tall person may be considered unusually tall, as they are 1.7 stdevs away from the mean height

  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.

It would be extreemly unlikely that the mean and stdev woulb be exactly the same as before, but I do think they would likely be close if the sample size was similar or larger.

  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 can use the standard error.

sdhgt/(length(bdims$hgt))**0.5
## [1] 0.4177887

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, becuase we know (100% sure) the mean of these adults is in that range. Were we to generalize to the population (all America adults), then the confidence interval would apply.

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

False, becuase the confidence interval can be used with non-normal distributions (although it may be better to calculate it a different way).

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

True

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

True: this is what CIs are for.

  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, a 90% confidence interval would be narrower as it can be less sure about the esitmate.

  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; it would need to be 9 times larger, as its related to the sqrt proportion of the sample size.

  1. The margin of error is 4.4.

True. Either end of the CI is 4.4 from the mean.

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. The observations random gifted children, the distribution is roughly normal, and the observations 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.

The null hypothesis here is when the average age at which gifted children first count to 10 sucessfully not less than 32 months; the alternative is that it is less than 32 months. The signifcance level is 0.10, so the P value needs to be less than 0.10. 

#Finding the probability of the null hypothesis
1-pnorm(32,30.69,4.31)
## [1] 0.3805852

The probability that the average age is not less than 32 months is 0.38, which is greater than our significance level of 0.10; thus we reject the null hypothesis.

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

As P value isnt in line with out significance level, we reject the null hypothesis: the children do learn to count faster than average children

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
#Standard Error
#A signifcance level of 10 corrisponds to a Z score of 1.645
SE = 1.645*4.31/(36)**0.5
SE
## [1] 1.181658
#Confidence Interval
CI = 30.69 + c(-SE,SE)
CI
## [1] 29.50834 31.87166

The confidence interval is 29.51 - 31.87

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

They do agree. The results suggest that the gifted children do learn to count to 10 significantly earlier than average children. The p value rejects the null hypothesis and the sample CI excludes the null hypothesis.

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.
#Probability of the averge IQ being over 100 
pnorm(100,118.2,6.5)
## [1] 0.00255513

The probability of the averge IQ of gifted clildren mothers being less than 100 (same as average mothers) is 0.0026, which is less than the significance of 0.1. Thus the mothers above are liekly to be different from average mothers.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
#Standard Error
#A signifcance level of 10 corrisponds to a Z score of 1.645
SE = 1.645*6.5/(36)**0.5
SE
## [1] 1.782083
#Confidence Interval
CI = 118.2 + c(-SE,SE)
CI
## [1] 116.4179 119.9821

The confidence interval is 116.42 - 119.98

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

They do agree. The probability of the mothers in question having an IQ less than or equal to 100 is is less than the significance level, and 100 falls well outside the confidence interval.

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 a mean is when you take random samples from a population, and a plot them. The distribution it forms will be normal, with a center around the population mean. The mean will remain around the popualtion mean even as sample size changes. The spread will tighten as the sample size increases. The shape will remain normal.

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?
1 - pnorm(10500,9000,1000)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.

The mean of this sampling distribution is the same as the population: 9000 hrs. The standard deviation of the samling distribution (standard error) is 1000/(15)^0.5 = 258.

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
1-pnorm(10500,9000,258)
## [1] 3.050719e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
library(ggfortify)
## Warning: package 'ggfortify' was built under R version 3.6.3
## Loading required package: ggplot2
## 
## Attaching package: 'ggplot2'
## The following object is masked from 'package:openintro':
## 
##     diamonds
library(ggplot2)


#Population
ggdistribution(dnorm,seq(5000,13000,100), mean = 9000, sd = 1000)

#Sampling
ggdistribution(dnorm,seq(5000,13000,100), mean = 9000, sd = 258)

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

No, becuase we need to assume normality in order to estimate probabilities here.

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 p-value will decrease, as it has an inverse correlation to sample size; the larger the sample size, the lower the p-value.