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.38 Height 150 160 170 180 190 200 0 20 40 60 80 100 Min 147.2 Q1 163.8 Median 170.3 Mean 171.1 SD 9.4 Q3 177.8 Max 198.1 (a) What is the point estimate for the average height of active individuals? What about the median?

Answer: Point estimate = Mean = 171.1 Median = 170.3

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

Answer: Point estimate of the SD = 9.4 IQR = Q3 - Q1 = 177.8 - 163.8 = 14

  1. 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.

Answer: Z = (180 - 171.1) / 9.4 = 0.95 Any Z score within -2 and 2 is within the normal limits and not high. Hence this value of 180 cms is not unusually high.

  1. 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.

Answer: The mean and standard deviation of each different sample selected will be different. Some samples might have more of high values taking the mean to a higher value. Others might have more of lower values taking the mean to a low value.

  1. 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.

Answer : variability of sample means = SD of any sample / sqrt(number in the sample) = 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.

  1. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11. Answer: False. Confidence level is a opulation parameter, and the saple suggests that 95% of the population is between $80.31 and $89.11.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. Answer: False. The sample size is 507, which is a big enough number to ignore the skew. And we can be lenient in this regards.

  3. 95% of random samples have a sample mean between $80.31 and $89.11. Answer: False. The confidence level is not about the sample mean, and is for the population.

  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. Answer: True. The confidence level is a prameter of the whole population, and suggests that percentage of the population will lie between this range.

  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. Answer: True. The lesser percentage of confidence means more values are outside the range, and hence narrower will be the confidence level.

  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. Answer: False. The margin of error = SD/sqrt(sample size). So if we want to reduce the margin of error by 3, the sample size must be 9 times larger, as sqrt(9) = 3, which will make the margin of error = 1/3 times

  7. The margin of error is 4.4. Answer: margin of error = half the width of 95% confidence level = (89.11 - 80.31) / 2 = 8.8/2 = 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.43

  1. Are conditions for inference satisfied? Answer: 3 conditions need to be justified for inference: Random - it is a random sample as mentioned Independence - As it is a simple random sample, it can be assumed to be independent Normal - There does not seem to a very high skew. So, this condition of normal distribution can also be assumed to be true So we can safely say that conditions for inference are satisfied.

  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 fist count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

Answer: H0 - denotes the Null Hypothesis H1 - denotes the Alternative Hypothesis H0 –> CHILDREN COUNT up to 10 at 32 months

Z score for H0 calculated below:

ZH0 <- (32-30.69) / (4.31/sqrt(36))
ZH0
## [1] 1.823666

Pvalue = 1 - P(ZHO) = 1 - 0.9656 = 0.0344

Pvalue < significance value for this case as 0.0344 < 0.1

We reject the Null Hypothesis whenever Pvalue < signifance value. Hence we can reject the null hypothesis - CHILDREN COUNT up to 10 at 32 months

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

Answer: We reject the Null Hypothesis whenever Pvalue < signifance value. Hence we rejected the Null Hypothesis here

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully. Answer : Z* from 90% Confidence Level = 1.645

SE = SD/sqrt(n)

SE <- 4.31 / sqrt(36)
Ninety_percent_CI_high_value <- 30.69 + 1.645 * SE
Ninety_percent_CI_high_value
## [1] 31.87166
Ninety_percent_CI_low_value <- 30.69 - 1.645 * SE
Ninety_percent_CI_low_value
## [1] 29.50834

Hence 90 percent CI is (29.50834, 31.87166)

  1. Do your results from the hypothesis test and the confidence interval agree? Explain. Answer: H0 was rejected in b. Also from (d), we see that 32 does not lie in 90 per cent CI. That means the rejection of Null Hypothesis goes well with the 90% confidence interval calculation. So both results match.

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.

  1. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average IQ of mothers of gifted children is di???erent than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

Answer: H0 = average IQ of mothers with gifted children equal to 100

null_value <- 100
mothers_sample_mean <- 118.2
SE_mothers <- 6.5 / sqrt(36)

mothers_Z_value <- (mothers_sample_mean - null_value) / SE_mothers

mothers p-value < 1 - 0.9998 and hence p-value < 0.0002 p-value < signifance level, hence we can safely reject H0 and say that the average IQ of mothers having gifted children is different than overall average IQ of 100.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children. Answer: For 90 % CI, Z* = 1.645

Hence,

Ninety_percent_CI_mothers_gited_hv <- 118.2 + 1.645 * SE_mothers
Ninety_percent_CI_mothers_gited_lv <- 118.2 - 1.645 * SE_mothers
sprintf("High value: %s", Ninety_percent_CI_mothers_gited_hv)
## [1] "High value: 119.982083333333"
sprintf("Low value: %s", Ninety_percent_CI_mothers_gited_lv)
## [1] "Low value: 116.417916666667"

Hence 90% CI is (116.4179, 119.9821)

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

Answer: Yes the answers match. In a, we rejected the H0 hypothesis which is average IQ of mothers with gifted children = 100 Also in b, we see the value 100 does not lie in 90% CI (116.4179, 119.9821). Hence both the answers match