Chapter 5 - Foundations for Inference

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.

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

Ans:

summary(bdims$hgt)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1

# the point estimate, which is the mean height:
mean(bdims$hgt)

## [1] 171.1438

# median:
median(bdims$hgt)

## [1] 170.3

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

Ans:
```
# point estimate of standard deviation:
s_d <- sd(bdims$hgt)
s_d
```
```
## [1] 9.407205
```
```
# IQR:
IQR(bdims$hgt)
```
```
## [1] 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.

Ans:

As the distribution is approximately normal, 180cm is above mean and 0.94 sd away from the mean, which is fairly tall.

A personal who is 155cm is below mean at 1.72 sd away from the mean, which unusually short.

hgt_mean <- mean(bdims$hgt)

# 1 sd range:
sd1 <- c(hgt_mean - s_d, hgt_mean + s_d)
sd1

## [1] 161.7366 180.5510

# 2 sd range:
sd2 <- c(hgt_mean - 2*s_d, hgt_mean + 2*s_d)
sd2

## [1] 152.3294 189.9582

#Using the formula: Z = (x - sample mean) / (sample sd)

# Z of 180cm
(180 - hgt_mean) / s_d

## [1] 0.9414287

# z of 155cm
(155 - hgt_mean) / s_d

## [1] -1.716109

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.

Ans:

The mean and standard deviation will be different from the ones given above. It is possible to get similar values but nearly impossible to get the exact same values because the sample size and values vary.
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 $SD_x = \frac{\sigma}{\sqrt{n}}$)? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Ans:

We use standard error to quantify the variability. The standard error of the original sample is around 0.418.
```
SDx <- s_d/sqrt(nrow(bdims))
SDx
```
```
## [1] 0.4177887
```

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.

Ans:

True, because we are given that the 95% confidence interval based on this sample is ($80.31, $89.11) hence we are confident that the average spending of all American adults are within that range.
This confidence interval is not valid since the distribution of spending in the sample is right skewed.

Ans:

False. The distribution is slightly right-skewed but the sample size is 436, which is fairly large.
95% of random samples have a sample mean between $80.31 and $89.11.

Ans:

False. Confidence interval is for population parameter, it cannot represents another sample. The confidence interval given in this question is from one sample and for population mean approximation. Other samples may have different 95% confidence intervals.
We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

Ans:

True. Same reason as part (b).
A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

Ans:

True. The range of 90% confidence interval is narrower than and within the 95% confidence interval. The smaller the percentage of confidence interval, the narrower the range is, noted that the sample mean is the same.
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.

Ans:

By using the standard error formula, $SE = \frac{\sigma}{\sqrt{n}}$, in order to have a new SE to a third of what is now, we would need to use a sample 9 times larger.

The margin of error is 4.4.

Ans:

True. The margin of error at 95% confidence interval = 1.96*SE = 4.405.

# 95% Confidence interval = (sample mean - 1.96*SE, sample mean + 1.96*SE)
# margin of error = critical value * standard error

n <- 436
SE <- sd(tgSpending$spending) / sqrt(n)
1.96 * SE

## [1] 4.405038

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?

Ans:

Yes, the observations are independent of each other and the distribution is not seriously skewed.

summary(gifted$count)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   21.00   28.00   31.00   30.69   34.25   39.00

library(moments)
skewness(gifted$count)

## [1] -0.1971696

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.

Ans:

Null Hypothesis H0: $\mu$ = 32 months

Alternative Hypothesis H1: $\mu$ < 32 months

As the p-value we got is lower than the significance level 0.1, we reject the null hypothesis.
```
#standard error
SE <- sd(gifted$count)/sqrt(nrow(gifted))
SE
```
```
## [1] 0.7191479
```
```
# Z-score: how many standard deviation from mean
(32 - mean(gifted$count))/SE
```
```
## [1] 1.81542
```
```
# the probability of getting our sample mean 30.69 months
pnorm(q=mean(gifted$count), mean=32, sd=SE, lower.tail = TRUE)
```
```
## [1] 0.03472969
```
Interpret the p-value in context of the hypothesis test and the data.

Ans:

P-value is the probability that we have 30.69 months as our sample mean with size 36 if the true mean is 32 months.

As the significance level $\alpha$ = 0.1 and the p-value we got is 0.035 < 0.1, we reject the null hypothesis.

Therefore, we strongly believe that gifted children fist count to 10 successfully is less than the general average of 32 months.
```
#standard error
SE <- sd(gifted$count)/sqrt(nrow(gifted))
SE
```
```
## [1] 0.7191479
```
```
# Z-score: how many standard deviation from mean
(mean(gifted$count)-32)/SE
```
```
## [1] -1.81542
```
```
# the probability of getting our sample mean 30.69 months
pnorm(q=mean(gifted$count), mean=32, sd=SE, lower.tail = TRUE)
```
```
## [1] 0.03472969
```
Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

Ans:

The 90% confidence interval is (29.51, 31.88).
```
# 90% confidence interval
c(mean(gifted$count)-1.645*SE, mean(gifted$count)+1.645*SE)
```
```
## [1] 29.51145 31.87744
```
Do your results from the hypothesis test and the confidence interval agree? Explain.

Ans:

Yes, I have the 90% confidence interval at (29.511, 31.877), which means there is 90% chance for the true mean falls in this range. The H0 with value 32 months falls out of the range which is very unlikely.

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.

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.

Ans:

Null Hypothesis H0: $\mu =$ 100

Alternative Hypothesis H1: $\mu \neq$ 100

As the p-value we got is 0, which is smaller than the significance level $\alpha$ = 0.1/2 = 0.05, we reject the null hypothesis.
```
# standard error: 1.083333
SE <- 6.5/sqrt(36)
SE
```
```
## [1] 1.083333
```
```
# Z-score: how many standard deviation from mean: 16.8
(118.2-100)/SE
```
```
## [1] 16.8
```
```
# the probability of getting our sample mean 118.2
1 - pnorm(q=118.2, mean=100, sd=1.083333, lower.tail = TRUE)
```
```
## [1] 0
```
Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

Ans:

The 90% confidence interval for the average IQ of mothers of gifted children is (116.42, 119.98).
```
# 90% confidence interval
c(118.2 - 1.645*SE, 118.2 + 1.645*SE)
```
```
## [1] 116.4179 119.9821
```
Do your results from the hypothesis test and the confidence interval agree? Explain.

Ans:

Yes, I have the 90% confidence interval at (116.42, 119.98), which means there is 90% chance for the true mean falls in this range. The rejected H0 with value 100 is far away from the CI.

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.

Ans:

“Sampling distribution” of the mean: it means taking n samples of sample size from the population and measure their means, and use the means to form a distribution.

When the sample size n increases, the shape of distribution becomes more normal, the center becomes closer to the true population mean, and the spread decreases.

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?

Ans:

The probability is 6.68%.
```
1 - pnorm(10500, mean=9000, sd=1000)
```
```
## [1] 0.0668072
```
Describe the distribution of the mean lifespan of 15 light bulbs.

Ans:

The distribution of 15 light bulbs would be nearly normal, with mean 9000 hours and sd 258.20 hours.
```
# sample standard deviation = population sd / sqrt(n)
SE <- 1000/sqrt(15)
SE
```
```
## [1] 258.1989
```
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

Ans:

The probability is 3.133e-09, which is about 0.
```
1 - pnorm(10500, 9000, 1000/sqrt(15))
```
```
## [1] 3.133452e-09
```

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

Ans:

x_lifespan <- 5200:12800
plot(x_lifespan, dnorm(x_lifespan, 9000, 1000), type = "l", col = "red", ylim = c(0,0.002))
lines(x_lifespan, dnorm(x_lifespan, 9000, SE), col = "blue" )
legend(10500, 0.0015, legend=c("Population", "Sample of size 15"),
      col=c("red", "blue"), lty=1:2, cex=0.8)

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

Ans:

Probability with a skewed distribution cannot be estimated.

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.

Ans:

As the margin of error = sd/sqrt(n), when the sample size increases, SE will decrease, then the |Z-score| (absolute value) will increase, and finally the p-value will decreases.

Chapter 5 - Foundations for Inference

Sin Ying Wong

10/12/2019

Heights of adults. (7.7, p. 260)

Thanksgiving spending, Part I.

Gifted children, Part I.

Gifted children, Part II.

CLT.

CFLBs.

Same observation, different sample size.