Data 606 Homework 4
Sergio Ortega Cruz
October 19, 2018
Title: “DATA 606 Homework 4”
date: “10/19/2018”
output: html_document
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.
library(DATA606)## Loading required package: shiny## Loading required package: openintro## Please visit openintro.org for free statistics materials##
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## demohist(bdims$hgt)
summary(bdims$hgt)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 147.2 163.8 170.3 171.1 177.8 198.1Min 147.2 Q1 163.8 Median 170.3 Mean 171.1 SD 9.4 Q3 177.8 Max 198.1
-(a) What is the point estimate for the average height of active individuals? What about the median?
point estimate for average height: 171.1 cm median: 170.3 cm-(b) What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
standard deviation: 9.4cm
IQR: 177.8cm ??? 163.8cm = 14cm-(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.
Both 180cm and 155cm are not unusual considering their Z scores are within 2 stardard deviations
```r
Z180 <- (180 - 171.1) / 9.4; Z180
```
```
## [1] 0.9468085
```
```r
Z155 <- (155 - 171.1) / 9.4; Z155
```
```
## [1] -1.712766
```
-(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.
No a random sample will included many but not all individuals that were not in the first sample, the second sample will have a different mean and standard deviation than the first sample. We can expect that the second mean and standard deviation will be close to the first.-(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¯=??n???SDx¯=??n)? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.
standard error which is the standard deviation divided by the square root of the sample size( nn).
SDx¯=9.4cm507???=0.4174687cmSDx¯=9.4cm507=0.4174687cm4.14 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,$89.11).
hist(tgSpending$spending)
summary(tgSpending$spending)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 5.719 49.180 75.790 84.710 112.300 282.800sd(tgSpending$spending)## [1] 46.92851Determine 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.
False, we are 95% confident that the mean spending for the entire population between $80.31 and $89.11-(b) This confidence interval is not valid since the distribution of spending in the sample is right skewed.
False, the validity of the confidence interval is based on the sample mean being Normally Distributed, not the sample itself. Since these people were selected at random and the sample size is greater than 30, by the Central Limit Theorem, the sample mean is normally distributed and the confidence intervals are valid.-(c) 95% of random samples have a sample mean between$80.31 and $89.11.
False, 95% of random samples produced confidence intervals that captures the population mean.-(d) We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
TRUE. The interval represents 2 standard deviation from point estimate, and approximates the average spending of the population-(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 the confidence interval, the less wide the net in approximation-(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, margin of error is z???xSEz???xSE, and SE goes as the inverse square root of sample. To decrease margin of error by 3 we need to increase sample size by 9 because 13=19???13=19.-(g) The margin of error is 4.4.
True, Margin of error is z???xSEz???xSE, which is 1/2 the confidence interval so, $89.11???$80.312=$4.44.24 Gifted children, Part I.
-Researchers investigating characteristics of gifted children collected data from schools in a large city on a random sample of thirty-six children who were identified as gifted children soon after they reached the age of four. The following histogram shows the distribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.
hist(gifted$count)
sd(gifted$count)## [1] 4.314887n = 36, min = 21, mean = 30.69, sd = 4.31, max = 39-a)Question: Are conditions for inference satisfied?
Yes, the children were randomly selected, independent of each other, and the sample size is greater than 30.
b)Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average age at which gifted children first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.
H0 = 32 months
HA < 32 months
??= 0.10c)Interpret the p-value in context of the hypothesis test and the data.
The small p-value of 0.034 suggests that there is sufficient evidence to reject H0 in favor of HA.
We reject the null hypothesis when the p-value is less than 0.10 in this caseZ_score <- (30.69 - 32) / (4.31 / sqrt(36)); Z_score
p_value <- pnorm(Z_score); p_value
d)Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
lo <- 30.69 - (qnorm(0.95) * (4.31 / sqrt(36))); lo## [1] 29.50845 up <- 30.69 + (qnorm(0.95) * (4.31 / sqrt(36))); up## [1] 31.87155 c(lo,up)## [1] 29.50845 31.87155e)Do your results from the hypothesis test and the confidence interval agree? Explain.
Yes, the hypothesis testing and confidence interval agree. Both support the alternative hypothesis in providing convincing evidence that the average age at which gifted children first count to 10 successfully can be less than the general average of 32 months4.26 Gifted children, Part II.
Exercise 4.24 describes a study on gifted children. In this study, along with variables on the children, the researchers also collected data on the mother’s and father’s IQ of the 36 randomly sampled gifted children. The histogram below shows the distribution of mother’s IQ. Also provided are some sample statistics. n 36 min 101 mean 118.2 sd 6.5 max 131
hist(gifted$motheriq)
summary(gifted$motheriq)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 101.0 113.8 118.0 118.2 122.2 131.0sd(gifted$motheriq)## [1] 6.504943-(a) Perform a hypothesis test to evaluate if these data provide convincing evidence that the average 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 = 100 IQ
HA ??? 100
?? = 0.10 Z_score <- (118.2 - 100) / (6.5 / sqrt(36)); Z_score## [1] 16.8 p_value <- 2 * (pnorm(Z_score, 0, 1, lower.tail = FALSE)); p_value## [1] 2.44044e-63p is much lower than 0.1 so we can reject the null hypothesis that the sample mean IQ is the same as the population mean IQ of 100 and take the alternative hypothesis that the sample mean IQ is different from the population mean IQ.- Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
lo <- 118.2 - (qnorm(0.95) * (6.5 / sqrt(36))); lo## [1] 116.4181up <- 118.2 + (qnorm(0.95) * (6.5 / sqrt(36))); up## [1] 119.9819c(lo,up)## [1] 116.4181 119.9819Do your results from the hypothesis test and the confidence interval agree? Explain.
They agree, and both support rejecting the null hypothesis. In the 2-sided hypothesis testing, the p-value returned is so small that it is too far less than the significance level of 0.10, while the confidence interval does not include the average IQ of 100 for the population.
4.34 Central Limit Theorem
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 distributions" is the distribution of sample means, which closely resembles the normal distribution. The normal model for the sample mean tends to be very good when the sample consists of at least 30 independent observations and the population data are not strongly skewed. The approximation can be poor if the sample size is small, but it improves with larger sample sizes4.40 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.
- What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
pnorm(10500, 9000, 1000, lower.tail = FALSE)## [1] 0.0668072b)Describe the distribution of the mean lifespan of 15 light bulbs.
since the population has normal distribution, the sample also would approximates normal distribution $N(9000, 1000/sqrt(15))$
A 6.68% chance that a single bulb will last 10,500 hours.- What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
pnorm(10500, 9000, 1000/sqrt(15), lower.tail = FALSE)## [1] 3.133452e-09d)Sketch the two distributions (population and sampling) on the same scale.
par(mfrow = c(2, 1))
xpop <- 7000:12000
ypop <- dnorm(xpop,mean=9000,sd=1000)
xsample <- 7000:12000
ysample <- dnorm(xsample,mean=9000,sd=1000/sqrt(15))
plot(xpop,ypop)
plot(xsample, ysample)
e)Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
It is likely not possible to get the estimates if the distribution is not somewhat normal, (or presumed normal). The calculation relies on conditions being met, one is that the distribution is not strongly skewed; if it is, then the sample should be sufficiently large4.48 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 decreases. As the sample increases, the standard error decreases which affects the z-score positively. When the area under the curve increases for the z-score, it decreases the complementary area under the tail of the distributions.