“Chapter 5 - Foundations for Inference”
“Hazal Gunduz”
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.
library(openintro)## Loading required package: airports## Loading required package: cherryblossom## Loading required package: usdatadata(bdims)
head(bdims)## # A tibble: 6 × 25
## bia_di bii_di bit_di che_de che_di elb_di wri_di kne_di ank_di sho_gi che_gi
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 42.9 26 31.5 17.7 28 13.1 10.4 18.8 14.1 106. 89.5
## 2 43.7 28.5 33.5 16.9 30.8 14 11.8 20.6 15.1 110. 97
## 3 40.1 28.2 33.3 20.9 31.7 13.9 10.9 19.7 14.1 115. 97.5
## 4 44.3 29.9 34 18.4 28.2 13.9 11.2 20.9 15 104. 97
## 5 42.5 29.9 34 21.5 29.4 15.2 11.6 20.7 14.9 108. 97.5
## 6 43.3 27 31.5 19.6 31.3 14 11.5 18.8 13.9 120. 99.9
## # … with 14 more variables: wai_gi <dbl>, nav_gi <dbl>, hip_gi <dbl>,
## # thi_gi <dbl>, bic_gi <dbl>, for_gi <dbl>, kne_gi <dbl>, cal_gi <dbl>,
## # ank_gi <dbl>, wri_gi <dbl>, age <int>, wgt <dbl>, hgt <dbl>, sex <int>(a) What is the point estimate for the average height of active individuals? What about the median?
x <- bdims$hgt
mean(x)## [1] 171.1438median(x)## [1] 170.3=> The point estimate for the average height of active individuals is 171.
=> And the median is 170.3
(b) What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
summary(x)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 147.2 163.8 170.3 171.1 177.8 198.1sd <- sd(x)
IQR(x)## [1] 14=> The IQR is 14.
(c) Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.
(180 - 171.1) / 9.4## [1] 0.9468085(155 - 171.1) / 9.4## [1] -1.712766=> A person who is 180cm is not considered unusually tall. A person who is 155cm tall is not considered unusually short. This is because they are only 1.71 below the standard deviation.
=> Z-score for 180 cm tall is 0.946; which is between -1 and 1 and its NOT unusual.
=> Z-score for 155 cm short is -1.716; which is NOT between -1 and 1 and its unusual.
(d) The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning.
hist(bdims$hgt, probability = TRUE)
x <- 140:200
y <- dnorm(x = x, mean = bdims$hgt, sd = bdims$hgt)=> It is unlikely that mean and standard deviation would be same, but it should be similar if the sample is a good representation of the population.
(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 SDx = √σ )? Compute this quantity using the n data from the original sample under the condition that the data are a simple random sample.
n <- length(bdims$hgt)
sd_x <- sd / sqrt(n)
sd_x## [1] 0.4177887=> Standard Error 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.
=> This is false; confidence interval is used for the population.
(b) This confidence interval is not valid since the distribution of spending in the sample is right skewed.
=> False because the sample size is 436 and it is large enough to not worry about the right skew.
(c) 95% of random samples have a sample mean between $80.31 and $89.11.
=> False because the confidence interval is used for the population mean, not the sample mean.
**(d*)** We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
=> True cause confidence interval is being used for the population mean.
(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, the lower confidence level, the narrower the interval gets.
(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, the sample size needs to be 9 times in this case.
(g) The margin of error is 4.4.
=> True.
89.11 - 84.71## [1] 4.4Gifted 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?
=> The conditions for inference are satisfied, sample size is 36 > 30, is randomly picked, and the results are not 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.
mean <- 30.69
n <- 36
sd <- 4.31
x <- 32
se <- sd / sqrt(n)
z_score <- (mean - x) / se
p <- pnorm(z_score, mean = 0, sd = 1)
p## [1] 0.0341013(c) Interpret the p-value in context of the hypothesis test and the data.
=> The above, the p-value is 0.0341.
(d) Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
lower <- mean - 1.645 * se
upper <- mean + 1.645 * se
c(lower, upper)## [1] 29.50834 31.87166(e) Do your results from the hypothesis test and the confidence interval agree? Explain.
=>Yes, average age is 32 months and the 90% confidence interval does not include 32 months.
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.
n 36 min 101 mean 118.2 sd 6.5 max 131
(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.
mean_2 <- 118.2
n_2 <- 36
sd_2 <- 6.5
x_2 <- 100
se_2 <- sd_2 / sqrt(n_2)
z_score_2 <- (mean_2 - x_2) / se_2
p_2 <- (1 - pnorm(z_score_2, mean = 0, sd = 1)) * 2
p_2## [1] 0=> The p value of 0 is not equal to 0.10; hence we reject the Null hypothesis (H0).
(b) Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
lower_2 <- mean_2 - 1.645 * se_2
upper_2 <- mean_2 + 1.645 * se_2
c(lower_2, upper_2)## [1] 116.4179 119.9821(c) Do your results from the hypothesis test and the confidence interval agree? Explain.
=> My results from the hypothesis test and the confidence interval do agree because the hypothesis test tells us to reject the null hypothesis that the average IQ of mother is 100 and the 90% confidence interval doesn’t include 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.
=> The term “sampling distribution” of the mean is the collection of means taken from multiple samples of the population. AS the sample size increaes, the shape becomes that of a normal distribution, the center becomes closer to the true population mean, and the spread starts to decrease.
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?
1 - pnorm(q = 10500, mean = 9000, sd = 1000)## [1] 0.0668072(b) Describe the distribution of the mean lifespan of 15 light bulbs.
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?
round(1 - pnorm(q = 10500, mean = 9000, sd = 258.20), 3)## [1] 0(d) Sketch the two distributions (population and sampling) on the same scale.
x <- 4000:14000
population <- dnorm(x, mean = 9000, sd = 1000)
sampling <- dnorm(x, mean = 9000, sd = 258.20)
plot(x, population, type = "l", col = "green")
lines(x, sampling, col = "blue")
(e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
=> You could not estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution because you would need a normal distribution and for part (c) the sample size is too small.
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 since you are increasing the sample size.
R-pubs => https://rpubs.com/gunduzhazal/816229