```
shhh <- suppressPackageStartupMessages # It's a library, so shhh!
suppressWarnings(shhh(library(ggplot2)))
suppressWarnings(shhh(library('DATA606'))) # Load the package
```

```
##
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics
## This package is designed to support this course. The text book used
## is OpenIntro Statistics, 3rd Edition. You can read this by typing
## vignette('os3') or visit www.OpenIntro.org.
##
## The getLabs() function will return a list of the labs available.
##
## The demo(package='DATA606') will list the demos that are available.
```

`suppressWarnings(shhh(library(knitr)))`

# 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.

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

Mean: 171.1

Median: 170.3

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

SD: 9.4

IQR: = Q3 - Q1

```
Q3 <- 177.8
Q1 <- 163.8
IQR <- Q3 - Q1
IQR
```

`## [1] 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:

```
x <- 180
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z
```

`## [1] 0.9468085`

z = 0.9468085 i.e less than 2, the person is not unusually tall.

155:

```
x <- 155
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z
```

`## [1] -1.712766`

z = -1.712766 i.e not beyond -2 the person is not unusually short.

## (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.

The new sample would be similar but not necessarily the same because if samples vary, their point estimates may also vary.

## (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

\(SD\bar { x } \quad =\quad \sqrt [ \sigma ]{ n }\) )?

##Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

```
sd <- 9.4
n <- sqrt (507)
sd/n
```

`## [1] 0.4174687`

# 4.14 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). 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.

FALSE. The confidence interval looks at the population not the sample size.

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

FALSE. The sample is sufficiently large (n = 436) to account for the skew.

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

FALSE. Confidence interval for the mean of a sample is not about other sample means.

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

TRUE. Its estimated by the point estimate and 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.

TRUE. With a 90% confidence interval we do not need such a wide interval to catch the values, so the interval would be narrower.

## (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. We will need 3 times large samples.

## (g) The margin of error is 4.4.

```
v1 <- 80.31
v2 <- 89.11
(v2 - v1)/2
```

`## [1] 4.4`

TRUE. The margin of error is half the confidence interval.

# 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.

## (a) Are conditions for inference satisfied?

Yes, The sample is random and 36 children of a large city is most likely under 10% of the population. The sample size is over 30. There doesn’t appear to be any strong skew in the population.

## (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.

```
x <- 32
n <- 36
avg <- 30.69
sd <- 4.31
se <- sd/sqrt(n)
z = (avg - x)/se
p = pnorm(z)
p
```

`## [1] 0.0341013`

0.0341013<0.10, we reject the null hypethesis H0 in favor of HA

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

If the null hypothesis is true, then the probability of observing a sample mean lower than 30.69 for a sample of 36 children is only 0.0341013 (p-value).

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

```
v1 <- avg - 1.65 * se
v2 <- avg + 1.65 * se
v1
```

`## [1] 29.50475`

`v2`

`## [1] 31.87525`

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

We are 90% confident that the average age at which gifted children first count to 10 is between 29.5 and 31.9 months.

# 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.

## (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.

```
x <- 100
n <- 36
avg <- 118.2
sd <- 6.5
se <- sd/sqrt(n)
z = (avg - x)/se
z
```

`## [1] 16.8`

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

```
v1 <- avg - 1.65 * se
v2 <- avg + 1.65 * se
v1
```

`## [1] 116.4125`

`v2`

`## [1] 119.9875`

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

Yes, We are 90% confident that the average IQ of mothers of gifted children is between 116.4125 and 119.9875. This is significantly above population average of 100.

## 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.

A sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. The sampling distribution of the mean takes the mean of each sample. As the sample size gets larger, the shape becomes on a more ‘normal’ and has a smaller spread.

# 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.

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

```
p <- 1 - pnorm(10500, 9000, 1000)
p
```

`## [1] 0.0668072`

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

The distribution of the mean lifespan of 15 light bulbs would be normal.

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

```
Z <- (10500 - 9000)/258
p <- 1 - pnorm(Z)
p
```

`## [1] 3.050719e-09`

The probability is 0%.

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

`normalPlot(9000, 1000)`

For 15 bulbs:

`normalPlot(9000, 1000/sqrt(15))`

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

No, We could not estimate parts (a) and (c) without a nearly normal distribution. For part (c) the sample size is too small.

# 4.48 Same observation, di???erent 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.

P-value will decrease as the sample size increases. With a smaller sd the Z-score will increase causing the p-value to decrease.

# Appendix

```
shhh <- suppressPackageStartupMessages # It's a library, so shhh!
suppressWarnings(shhh(library(ggplot2)))
suppressWarnings(shhh(library('DATA606'))) # Load the package
suppressWarnings(shhh(library(knitr)))
Q3 <- 177.8
Q1 <- 163.8
IQR <- Q3 - Q1
IQR
x <- 180
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z
x <- 155
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z
sd <- 9.4
n <- sqrt (507)
sd/n
v1 <- 80.31
v2 <- 89.11
(v2 - v1)/2
x <- 32
n <- 36
avg <- 30.69
sd <- 4.31
se <- sd/sqrt(n)
z = (avg - x)/se
p = pnorm(z)
p
v1 <- avg - 1.65 * se
v2 <- avg + 1.65 * se
v1
v2
x <- 100
n <- 36
avg <- 118.2
sd <- 6.5
se <- sd/sqrt(n)
z = (avg - x)/se
z
v1 <- avg - 1.65 * se
v2 <- avg + 1.65 * se
v1
v2
p <- 1 - pnorm(10500, 9000, 1000)
p
Z <- (10500 - 9000)/258
p <- 1 - pnorm(Z)
p
normalPlot(9000, 1000)
normalPlot(9000, 1000/sqrt(15))
``
```