Chapter 4 Foundations for Inference Practice: 4.3, 4.13, 4.23, 4.25, 4.39, 4.47 Graded: 4.4, 4.14, 4.24, 4.26, 4.34, 4.40, 4.48

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

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

Mean = 171.1 Medial = 170.3

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

SD=9.4

IQR = 177.8 - 163.8 = 14

- 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.

Range of 2 standard deviations of the mean is between:

171.1-2*9.4 = 152.3

171.1+2*9.4 = 189.9

Both of these observation are within this range so they are not considered unusual.

- 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 mean and standrand deviation will not be exactly the same but it will be similar. Point estimates that are based on samples only approximate the population parameter, and they vary from one sample to another.

- 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 ̄ = pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Let’s calculate the standard error for our sample: SE = 9.4/sqrt(507) = 0.42

4.14 Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked o↵ 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.

- We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

False. the sample mean is in the confidence interval, the 95% confidence applies to the population mean.

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

False. We can use condifence interval in this case - the data distribution is not very strongly skewed.

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

False. We are only 95% confidence that is the case for samples of the same size (n=436)

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

True.

- 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.

- 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 would need to be 9 times larger.

- The margin of error is 4.4.

True. (89.11 - 80.31)/2 = 4.4.

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.

- Are conditions for inference satisfied?

Yes. Here are the reasons: (1) Kids were randomly chosen; (2) The sample size is more than 30. (3) The distribution apears to be close enough to normal

- 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.

H0 = 32

H1 < 32

(30.69-32)/(4.31/sqrt(36)) = -1.823666

`pnorm(-1.823666)`

`## [1] 0.03410129`

Since the above valie is under 0.1 we reject the null hypothesis.

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

We conclude that the age at which gifted children count to 10 is lower than average.

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

`qnorm(0.05, lower.tail=FALSE)`

`## [1] 1.644854`

`30.69 - 1.645 * 4.31 / sqrt(36) `

`## [1] 29.50834`

`30.69 + 1.645 * 4.31 / sqrt(36)`

`## [1] 31.87166`

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

Yes, since our 90% confidence interval doesn’t include our null hypothesis mean of 32 whcih we rejected.

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.

- 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: IQ = 100

H1: IQ > 100

(118.2-100)/(6.5/sqrt(36)) = 16.8

`pnorm(16.8, lower.tail = FALSE)`

`## [1] 1.22022e-63`

Since O is lower than 0.1 - we reject our null hypothesis that the average IQ of mothers of gifted children is equal to average IQ.

- Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

`qnorm(0.05, lower.tail=FALSE)`

`## [1] 1.644854`

`118.2 - 1.645 * 6.5 / sqrt(36) `

`## [1] 116.4179`

`118.2 + 1.645 * 6.5 / sqrt(36)`

`## [1] 119.9821`

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

Yes, the 90% confidence interval does not include our H0 mean 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.

The sampling distribution represents the distribution of the point estimates based on samples of a fixed size from a certain population. As the sample size increases the shape, center and spread will get closer to the true population parameters (shape, mean and 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.

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

z = (10500 - 9000) / 1000 = 1.5

```
#The probability is:
round(pnorm(1.5, lower.tail = FALSE),3)
```

`## [1] 0.067`

- Describe the distribution of the mean lifespan of 15 light bulbs.

The distribution will be normal and will be centered around the population mean of 9000.

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

SE = 1000/sqrt(15) = 258.1989

z = (10500 - 9000)/ 258.1989 = 5.81

```
#The probability is:
round(pnorm(5.81, lower.tail = FALSE),3)
```

`## [1] 0`

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

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

Not if the distribution was strongly skewed.

4.48 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.

As sample size increases that will cause our SE to decrease. If SE is lower that would mean that Z is now higher. If Z increases the probability decreases so p-value will decrease.