Homework- Chapter 4

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.

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1. What is the point estimate for the average height of active individuals? What about the median? (See the next page for parts (b)-(e).)

• The point estimate for average height of active individuals is 171.1. Median is 170.3.

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

• The point estimate for standard deviation is 9.4. The IQR is 177.8-163.8= 14cm.

3. 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.

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

## [1] 0.9468085

Since z = 0.9468085 is < 2, 180 cm is not considered unusual.

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

## [1] -1.712766

Since z = -1.71 is < 2, 155 cm is not considered unusual.

4. 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 results of the new sample would not be identical. The results of the new sample would be similar due to the samples represent the population.

5. 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 \({ SD }_{ \overline { x } }=\frac { \sigma }{ \sqrt { n } }\) )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

\(SE=\frac { \sigma }{ \sqrt { n } } =\frac { 9.4 }{ 22.51666 } =0.42\)

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.

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1. 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 always in the confidence level.

2. This confidence level is not valid since the distribution of spending in the sample is right skewed.

False - the confidence interval still can be valid even if the distribution is right or left skewed.

3. 95% of random samples have a sample mean between $80.31 and $89.11

False- sample of different size may have different confidence level.

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

Yes, that’s true. The average spending of an average American adult is within the confidence interval with 95% of probability.

5. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

Yes, that’s True.

6. 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,

\(SE=\frac { \sigma }{ \sqrt { n } }\) , so 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.

7. The margin of error is 4.4.

(89.11-80.31)/2

## [1] 4.4

True, The margin of error is 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.

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1. Are conditions for inference satisfied?

Yes, conditions for inference satisfied: the children were randomly selected and independent of each other. The sample size is greater than 30, which is large enough. The graph is normally distributed and not obviously skewed.

2. 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.

\({ H }_{ 0 }\): \(\mu\)= 32 months.

\({ H }_{ A }\): \(\mu \neq32\) months.

\({ SE }_{ \overline { x } }=\frac { { s }_{ x } }{ \sqrt { n } } =\frac { 4.31 }{ \sqrt { 36 } }\)

##standard error

SE<- 4.31/sqrt(36)
SE

## [1] 0.7183333

\(Z=\frac { \overline { x } -null\quad value }{ SE_{ \overline { x } } } =\frac { 30.69-32 }{ 0.718 }\)

Z_score <- (30.69-32)/SE

Z_score

## [1] -1.823666

p_value <- pnorm(Z_score); 
p_value

## [1] 0.0341013

p-value = 0.0341 < 0.1, we reject the null hypothesis.

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

lowerTail = 30.69 -   1.645 * SE
upperTail = 30.69 +   1.645 * SE
lowerTail

## [1] 29.50834

upperTail

## [1] 31.87166

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

Both results agree. Hypothesis test proves the alternative hypothesis (\({ H }_{ A }\): \(\mu \neq32\) months), as well as confidence interval shows that the mean age wouldn’t be more than 31.87.

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.

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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

\({ H }_{ 0 }\): \(\mu\)= 100.

\({ H }_{ A }\): \(\mu \neq100\)

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-63

lowerTail = 118.2 -   1.645 * (6.5 / sqrt(36))
upperTail = 118.2 +   1.645 * (6.5 / sqrt(36))
lowerTail

## [1] 116.4179

upperTail

## [1] 119.9821

The 90% Confidence interval is from 116.42 to 119.98

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

Yes, agree and both support the rejecting of null hypothesis.

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 distribution is the probability distribution of the sample from the population.

As sample sizes increase, the sampling distributions approach a normal distribution. As sample sizes increase, the variability of each sampling distribution decreases.

4.40 CFLBs.

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

 1 - pnorm(10500,mean=9000,sd=1000)

## [1] 0.0668072

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

Due to the population has normal distribution, the sample also would approximates normal distribution.

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

1-pnorm(10500,mean=9000,sd=258)

## [1] 3.050719e-09

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

xpopulation <- 7000:12000
ypopulation <- dnorm(xpopulation,mean=9000,sd=1000)

xsample <- 7000:12000
ysample <- dnorm(xsample,mean=9000,sd=1000/sqrt(15))

plot(xpopulation,ypopulation)

plot(xsample, ysample)

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 the sample size increases, p-value decreases.

As sample sizes increase, the variability of each sampling distribution decreases as well as p-value gets smaller.