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

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? The point estimate for the average height of active individuals is 171.1 cms. The point estimate for the median height of active individuals is 170.3 cms.

What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR? The point estimate for the standard deviation of the heights of active individuals is 9.4 cms. The point estimate for the IQR of the heights of active individuals is 14 cms.

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.

`require(openintro)`

`## Loading required package: openintro`

`## Please visit openintro.org for free statistics materials`

```
##
## Attaching package: 'openintro'
```

```
## The following objects are masked from 'package:datasets':
##
## cars, trees
```

`mean(bdims$hgt)`

`## [1] 171.1438`

```
mu<-171.1
sigma<-9.4
(z180<-(180-mu)/sigma)
```

`## [1] 0.9468085`

`1-pnorm(z180 )`

`## [1] 0.1718682`

`(z155<-(155-mu)/sigma)`

`## [1] -1.712766`

`pnorm(z155)`

`## [1] 0.0433778`

```
#(ci95.up <- mu+1.96*stderror)
#(ci95.dn <- mu-1.96*stderror)
```

Since z180 is less than 1 standard deviation above from the mean, the person who is 180 cm would not be considered unusual. About 17.18% of the population is taller than 180 cm.

Since z155 is about 1.71 standard deviations below from the mean, the person who is 155 cm could be considered unusually short. Only 4.33% of the population is shorter than 155 cm.

- 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, the mean and standard deviation of this new sample will almost certainly be different from the previous sample, because the samples are being taken randomly and the population of physically active individuals is much larger than the sample size of 507.

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

`(stderror<-9.4/sqrt(507))`

`## [1] 0.4174687`

The standard error quantifies the variability of the sample mean. The standard error is 0.417.

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.

We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11. False. The average spending for these 436 adults is already given as the point estimate.

This confidence interval is not valid since the distribution of spending in the sample is right skewed. False. There is no information provided about the skewness of the underlying population or the sample.

95% of random samples have a sample mean between $80.31 and $89.11. False. The interval above is expected to contain the population mean.

We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. True. This is the interval for the average daily spending of all American adults, that can be inferred based on the sample mean, sample size and the chosen confidence interval.

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 than the 95% confidence interval, since it “enables” us to be less confident of our estimate.

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. 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 9 times larger.

The margin of error is 4.4. True. The 2 end-points of the interval are $4.4 away from the mean.

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, the sample size is greater than 30, and at the same time it is almost certain to be less than 10% of the population of the 4 year old children in the large city’s schools. Also, it is stated that the sample was collected randomly, so the observations should be independent.

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.

`(mean.G<-mean(gifted$count))`

`## [1] 30.69444`

`(sd.G<-sd(gifted$count))`

`## [1] 4.314887`

`(n.G<-nrow(gifted))`

`## [1] 36`

```
null.value.G<-32
z90<-1.285
(stderror.G<-sd.G/sqrt(n.G))
```

`## [1] 0.7191479`

`(margin.error.G<-z90*stderror.G)`

`## [1] 0.9241051`

`(lower.cutoff<-32-margin.error.G)`

`## [1] 31.07589`

`(z.G<-(mean.G - null.value.G)/stderror.G)`

`## [1] -1.81542`

`(pnorm(z.G))`

`## [1] 0.03472969`

H0 = Average age at which gifted children first count to 10 is 32 months or higher Ha = Average age at which gifted children first count to 10 is less than 32 months Sample size = 36 children Sample mean = 30.69 months Sample sd = 4.31 months

Based on the fact that the p-value of 3.47% is lower than the significance level of 10%, the Null Hypothesis can be rejected, in favor of the Alternative Hypothesis. The same can be seen from the fact that the lower cutoff (left tail) is 31.07 based on a null value of 32 and a margin of error = 0.924. Since the sample mean at 30.69 is lower than this lower cutoff, it implies that the Alternative Hypothesis is to be accepted.

Interpret the p-value in context of the hypothesis test and the data. p-value = probability of observing data at least as favorable to Ha as our current data set (a sample mean = 30.69), if in fact H0 were true (the true population mean was 32). The low probability of 3.47% indicates that the probability of observing a sample mean of 30.69 is very low if the true population mean is indeed 32 months or higher.

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

`(conf.interval.lower.G<-mean.G-1.645*stderror.G)`

`## [1] 29.51145`

`(conf.interval.upper.G<-mean.G+1.645*stderror.G)`

`## [1] 31.87744`

The 90% confidence interval is (29.51, 31.87)

- Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, the results agree. The confidence interval shows that the null value is not included in the range around the sample mean, and the hypothesis test shows that the probability of observing the sample value is lower than the significance required.

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.

Mother’s IQ n 36 min 101 mean 118.2 sd 6.5 max 131

- 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 = Average IQ of mothers of gifted children = 100 Ha = Average IQ of mothers of gifted children <> 100 Sample size = 36 children Sample mean = 118.2 Sample sd = 6.5

`(mean.MIQ<-mean(gifted$motheriq))`

`## [1] 118.1667`

`(sd.MIQ<-sd(gifted$motheriq))`

`## [1] 6.504943`

`(n.MIQ<-nrow(gifted))`

`## [1] 36`

```
null.value.MIQ<-100
z90<-1.645
(stderror.MIQ<-sd.MIQ/sqrt(n.MIQ))
```

`## [1] 1.084157`

`(margin.error.MIQ<-z90*stderror.MIQ)`

`## [1] 1.783439`

`(lower.cutoff.MIQ<-100-margin.error.MIQ)`

`## [1] 98.21656`

`(upper.cutoff.MIQ<-100+margin.error.MIQ)`

`## [1] 101.7834`

`(z.MIQ<-(mean.MIQ - null.value.MIQ)/stderror.MIQ)`

`## [1] 16.75649`

`2*pnorm(-abs(z.MIQ))`

`## [1] 5.077477e-63`

Based on the range(98.21,101.78), it is clear that a sample mean of 118.16 is significantly different from the null value of 100, and so the null hypothesis can be rejected in favor of the alternative hypothesis. The p-value associated with this z-score is tiny, and below the significance.

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

`(conf.interval.lower.MIQ<-mean.MIQ-1.645*stderror.MIQ)`

`## [1] 116.3832`

`(conf.interval.upper.MIQ<-mean.MIQ+1.645*stderror.MIQ)`

`## [1] 119.9501`

The 90% confidence interval for the average IQ of mothers of gifted chikdren is (116.38, 119.95)

- Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, the results agree. The confidence interval shows that the null value is not included in the range around the sample mean, and the hypothesis test shows that the probability of observing the sample value is lower than the significance required.

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 of the mean is the distribution of means from randomly selected samples from the population of a given sample size. If the sample size is less than 10% of the population size, but greater than 30, then the sampling distribution is normal centered at the population mean, with a spread equal to the standard error (sample standard deviation divided by the square root of the sample size). As the sample size increases, its center remains the same, but its spread decreases.

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<-((10500-9000)/1000))`

`## [1] 1.5`

`(prob.10500<-1-pnorm(z.10500))`

`## [1] 0.0668072`

The probability that a randomly chosen light bulb lasts more than 10,500 hours is 6.68%

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

`(std.error.15bulbs = 1000/sqrt(15))`

`## [1] 258.1989`

The distribution of the mean lifespan of 15 light bulbs is normal with mean = 9000 hours and standard deviation = 258.19 hours.

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

`(z.15bulbs<-(10500-9000)/std.error.15bulbs)`

`## [1] 5.809475`

`(prob.15bulbs<-1-pnorm(z.15bulbs))`

`## [1] 3.133452e-09`

The probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours is almost 0.

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

`library(ggplot2)`

`## Warning: package 'ggplot2' was built under R version 3.5.3`

```
##
## Attaching package: 'ggplot2'
```

```
## The following object is masked from 'package:openintro':
##
## diamonds
```

```
#df.bulb.lifespan<-data.frame(popl<-rnorm(100,9000,1000), samp<-rnorm(100,9000,258.19))
curve(dnorm(x,9000,1000), xlim=c(5000, 13000), main="Population Bulb Lifespan", col="blue")
```

`curve(dnorm(x,9000,258.19), xlim=c(5000, 13000), main="Sample Bulb Lifespan", col="red")`

- Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? If the bulb lifespan had been skewed, then a sample size of 15 would not be sufficient.

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. When the sample size increases, the standard error and the margin of error decreases. This has the effect of narrowing the confidence interval. The z-value associated with the sample estimate would increase. As a result, the p-value would decrease.