606 Chap4 Assignment
Hui Gracie Han
November 12, 2018
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
- What is the point estimate for the average height of active individuals? What about the median?
#according to the figures and discription of the data
adult_mean <- 171.1
adult_median <- 170.3
paste("average height is: ", adult_mean)## [1] "average height is: 171.1"paste("median height is: ", adult_median)## [1] "median height is: 170.3"(b)What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
adult_sd <- 9.4
Q3 <- 177.8
Q1 <- 163.8
adult_IQR <- Q3 - Q1
paste("standard deviation is", adult_sd)## [1] "standard deviation is 9.4"paste("IRQ is", adult_IQR)## [1] "IRQ is 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.
#z value more or less than 2 sd would be considered as tall or short, respectively
adult_sd <- 9.4
adult_mean <- 171.1
z <- (180 - adult_mean)/2*adult_sd
if (z > 1) {
print("180 is considered tall")
} else {
print("180 is not considered tall")
} ## [1] "180 is considered tall"adult_sd <- 9.4
adult_mean <- 171.1
z <- (adult_mean -155)/(2*adult_sd)
if (z > 1) {
print("155 is considered short")
} else {
print("155 is not considered short")
} ## [1] "155 is not considered short"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(d): No, I would not expect the mean and the standard deviation between the two groups to be the same. Because the individual is randomly picked, the individual observations would not be the same. But it would expect them to be similar.
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? = p! n )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.
SE <- adult_sd/sqrt(507)
paste("Standard error is: ", round(SE,3))## [1] "Standard error is: 0.417" library(DATA606)## Loading required package: shiny## Loading required package: openintro## Please visit openintro.org for free statistics materials##
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## demo getwd()## [1] "C:/Users/graci/Documents/606 GracieHan 2018"#install.packages(DATA606)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, we cannot make inference on the population parameter, not on the point estimate. The point estimate is always in the confidence interval.
This confidence interval is not valid since the distribution of spending in the sample is right skewed. #False, If the sample size was lower than 30, this would be a problem; however our sample size is 436.
95% of random samples have a sample mean between $80.31 and $89.11. #False, the confidence interval is not about a sample mean.
We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. #True, this is the definition of a 95% 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
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, we would need to make our sample 9 times larger in order to have a confidence interval that is a third of what it is now. The reason is that we would need the square root of our sample size to be 3 times it’s current value in order to triple the standard error.
The margin of error is 4.4. #True, the margin of error is 4.4 since our confidence interval is equal to our point estimate plus and minus the margin of error. (89.11 - 84.71) = 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
Are conditions for inference satisfied? #Yes, (1) samples are random selected. (2) observation is independent. (3) less than 10% of population. (4) size more than 30. (5)no obvious skew on histogram.
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: average age of gifted children first time count 10 = average age of general children first time count 10 (32 months) HA: average age of girfted children first time count 10 < average age of general children first time count 10 (32 months)
z_score= (30.69 - 32) / 4.31
z_score## [1] -0.3039443library(DATA606)normalPlot(mean = 0, sd = 1, bounds = c(-Inf,z_score), tails = FALSE)
- Interpret the p-value in context of the hypothesis test and the data.
Because the p value is 0.381 which is much bigger than 0.1, there is no statistical significance between gifted children and general children population on first time counting 10.
- Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
low_tail <- round(30.69 - 1.65*4.31, 2)
up_tail <- round(30.69 + 1.65*4.31, 2)
low_tail## [1] 23.58up_tail## [1] 37.8- Do your results from the hypothesis test and the confidence interval agree? Explain.