4.4) a) point estimate for average height = 171.1 cm
point estimate for median height = 170.3 cm

b)point estimate for standard deviation = 9.4 cm
point estimate for IQR = 14 cm

c) To determine the range of heights that fall within 95% of the population (2 standard deviations away from the mean):
171.1 +/- 2(9.4)
152.3 cm and 189.9 cm
180 cm is within that range so that is not unusually tall
155 cm is within that range so that is not unusually short

d) No, I would not expect the mean and standard deviation of the next sample to be identical to the first. Each sample will vary. Point estimates are based on the sample and approximate the population parameters.

e) SE = \(\sigma\)/\(\sqrt{n}\)

SE <- 9.4/sqrt(507)
SE
## [1] 0.4174687

4.14)
a) False. Inference is made on the population parameter, not the point estimate.
b) False. The larger the sample, the more lenient you can be with the skew and this sample is sufficiently large.
c) True. The confidence interval means that 95% of the intervals contain the mean.
d) True. For the population, we would expect 95% of samples to be within the confidence interval.
e) True. If the confidence interval were lower, then we are less sure about our estimate.
f) False. The margin of error is proportional to 1 over the square root of the number of trials. If you want the margin of error to go down by 1/3, the number of samples needs to be increased by a factor of 9.
g) True. The margin of error is the difference between the mean and values that bound the confidence interval.

4.24)
a) conditions for inference: The sample size is larger than 30. The sample consists of less than 10% of the population of gifted children and the children were chosen randomly so the observations are independent. The sample observations should be independent and the distribution is fairly symmetrical. The conditions for inference are met.
b) Ho: \(\mu\) = 32 months The average age for children to learn to count to 10 is 32 months
HA: \(\mu\) < 32 months The average age for gifted children to learn to count to 10 is earlier than 32 months
SE = \(\sigma\)/\(\sqrt{n}\) 

SE <- 4.31/sqrt(36)
SE
## [1] 0.7183333
Z <- (30.69-32)/SE
Z
## [1] -1.823666

From looking at the normal table, p = 0.0344
c) The p value is less than the signifcance value. Therefore the data provides convincing evidence in favor of the alternate hypothesis. That is to say, it is likely that gifted children on average learn to count to 10 at an age younger than 32 months.
d) 90% confidence is associated with a Z score of 1.64
30.69 +- (1.64)(.72)
confidence interval = (29.5092,31.8708)
e) 90% of gifted children learn to count to 10 between the ages of 29.5 months and 31.9 months. This is younger than the age of 32 months to learn to count to 10 so the confidence interval and p score yield the same result.
4.26)
a) Ho: \(\mu\) = 100 The average IQ of the mothers of gifted children is 100.
HA: \(\mu\) \(\neq\) 100 The average IQ of the mothers of gifted children is not 100.
SE = \(\sigma\)/\(\sqrt{n}\) 

SE <- 6.5/sqrt(36)
SE
## [1] 1.083333
Z <- (118.2-100)/SE
Z
## [1] 16.8

The probability of a mean of 118.2 is less than .0002
Therefore the p score is less than .0004
The p score is much less than the significance level of .10. Therefore we can reject the null hypthesis. The data provides evidence for the alternate hypothesis. That is to say, the data supports the hypothesis that the average IQ of the mothers of gifted children is not 100.
b) 10% confidence interval 118.2 +- (1.64)(1.08)
confidence interval = (116.4288,119.9712)
c) The results from the hypothesis test and confidence interval agree. 90% of mothers of gifted children have an IQ between 116.4 and 120. Therefore the null hypothesis, which states that mothers have an average IQ of 100, should be rejected.
4.34)
The sampling distribution of the mean is comprised of the mean of many subsets of the data. As the sample size increases, the center approaches the mean of the population, the spread decreases and the shape becomes more normal.

4.40)
\(\mu\) = 9000 h
\(\sigma\) = 1000 h 

Z <- (10500-9000)/1000
Z
## [1] 1.5

The proportion of observations less than 10,500 hours is .9332
1 - .9332 = .0668
The probability a randomly chosen bulb has a life greater than 10,500 hours is .0668.
b) The distribution of the mean lifespan of 15 lightbulbs will be narrower than the distribution for larger number of light bulbs.
c)

SE <- 1000/sqrt(15)
SE
## [1] 258.1989
Zfifteen <- (10500-9000)/SE
Zfifteen
## [1] 5.809475

It is very unlikely that a light bulb will have a lifetime greater than 10,500 hours in a sample of just 15 bulbs. The probability is less than 0.0002.
d)

x <- 6000:12000
a <- dnorm (x=x, mean = 9000, sd=1000)
fifteen <- dnorm (x=x, mean = 9000, sd=1000/sqrt(15))
plot(x=x, y=fifteen, col = "red", xlab="number of hours", ylab="")
lines(x=x, y=a, col = "blue")


e) For part a, if the number of samples is very large, then you could still estimate the probabilities of a bulb having a lifespan greater than 10500 hours. However in part c, there were only 15 light bulbs. That is not a large enough sample if the sample is skewed.
4.48) p decreases with sample size Mathematically we can see this because the standard error depends on the sample size. If the sample size increases by a factor of 100, the standard error decreases by a factor of 10. This causes the Z score to increase by a factor of 10. This causes p to decrease.