4.4 Heights of adults.

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

The mean is 171.1. The median is 170.3

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

SD: 9.4 ; IQR: 163.8 - 177.8

(c) 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.

No, the person is tall but usually as 180 is not far from the center of the distribution.
155cm should be consider as unusually short as it situate at the left tail of the distribution.

(d) 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.

Not the same but should be similar. Since the samples are choosen randomly, the individual observations would not be the same. However, the sample should represent the population to some extend. Thus should be similar.

(e)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¯=σn√)? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Use Standard error.

sd = 9.4 / sqrt(507)
sd
## [1] 0.4174687

The SD should be 0.4174687

Exercise 4.14 Thanksgiving spending, Part I

(a) We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

False. We are 95% confident that the average spending of ALL American adults is between $80.31 and $89.11. We are 100% confident that the average spending of these 436 American adults (we already know) is between $80.31 and $89.11. #### (b) This confidence interval is not valid since the distribution of spending in the sample is right skewed. True. The confidence interval is not valid due to strong right skew.

(c) 95% of random samples have a sample mean between $80.31 and $89.11.

False, we cannot use this confidence interval to make any assumptions about other sample means when we don’t even know the size of other random samples.

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

True, this is the right interpretation of 95% confidence interval

(e) 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, 90% confidence is less certain than 95% confidence when the interval is narrower.

(f) 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 need to make sample size 9 times larger in order to decrease the margin of error to a third. (SE = sigma / sqrt(n))

(g) The margin of error is 4.4.

True

upper = 89.11 
lower = 80.31 
margin_error =(upper-lower)/2
margin_error
## [1] 4.4

Exercise 4.24 Gifted children, Part I

(a) Are conditions for inference satisfied?

Yes 1. sample size > 30
2. sample obersvations are independent with each other
3. distribution not strongly skewed
4. Less than 10% of the population

(b) 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.

H0: The average age of gifted children equals to the general population, average_age =32 months. HA: The average age of gifted children less than the general populaton, average_age <32 months.

z = (30.69 - 32) / 4.31
normalPlot(sd = 1,mean = 0, tails = FALSE  ,bounds = c(-Inf,z) )  ## one tail

(c) Interpret the p-value in context of the hypothesis test and the data.

Since the p value is 0.381, which is greater than the 0.1 significance level, we cannot reject the null hypothesis. Thus the average age of gifted children is not significantly lower than the general population.

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

mean = 30.69
sd = 4.31
sigma = 1.65
lower = mean-sigma*sd
upper = mean+sigma*sd
c(lower,upper)
## [1] 23.5785 37.8015

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

The results agrees. The general population average age 32 months falls in the 90% confidence interval 23.5785, 37.8015

Exercise 4.26 Gifted children, Part II

(a) 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: The average IQ of gifted children’s mother equals to the general population’ mother IQ, average_IQ =100. HA: The average IQ of gifted children’s mother is different than the general population’ mother IQ, average_IQ != 100.

z = (118.2 - 100) / 6.5
z
## [1] 2.8
normalPlot(mean = 0, sd = 1, bounds = c(-z,z), tails = TRUE) ##double tail

p = 1-0.995 
p
## [1] 0.005

Since the p-value is 0.005, which is smaller than the 0.1 significant level, we can draw the conclusion that the gifted mother’s average IQ is different than the general populations’ mothers’ average IQ.

(b) Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

mean = 118.2
sd = 6.5
sigma = 1.65
lower = mean-sigma*sd
upper = mean+sigma*sd
c(lower,upper)
## [1] 107.475 128.925

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

The results agree because the average IQ of general populations’ mother did not fall within the 90% confidence interval.

Exercise 4.34 CLT