4.4 Heights of adults.

(a) What is the point estimate and average height of active individuals? Median?

Point estimate is the sample mean, 171.1….Median is 170.3

(b) What is the Point Estimate for the SE of the heights of active individuals? IQR?

SD is 9.4….IQR is Q3 - Q1 so… 14 <<< Calculations can be seen in R markdown for this answer and future.

(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.

For different answers, lets define unusual as within 1 standard deviation. If so then our range of usual is 161.7 - 180.5.
180cm is not unusual.
155cm is unusual.

(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.

I would expect the mean and SD of the new sample to be similar, but not the same. If the individuals surveyed were selected with current scientific standards then any sample over 30 of these individuals should resemble a very similar distribution with roughly the same numbers. (of course there is always room for error!)

(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 SD¯ x = pn)? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

SE = sSD/sqrt(sampleSize)…0.4174687

4.14 Thanksgiving spending, Part I. TRUE OR FALSE w/ Reasoning

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

FALSE : We are 100% confident that the average spending “OF THESE 436 AMERICAN ADULTS” is between 80-89. Although these 436 Americans are our sample of America, we are making a statement about the sample, making them the population.

(b) 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.

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

FALSE : The CI is constructed from our sample mean, we can not induce the same interval from a different sample mean.

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

TRUE : This is what confidence intervals are supposed to tell us in the first place.

(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 : A 90% confidence interval is narrower, because when we do not need to be as confident in our answer, we can afford to be a bit more specific.

(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 have to remember that the formula calls for the square root of the sample size, therefore 3^2

(g) The margin of error is 4.4.

TRUE : ME = upper tail - lower / 2 … 4.4 = 89.11 - 80.31 / 2

4.24 Gifted children, Part I.

(a) Are conditions for inference satis???ed?

Yes
####(b) Suppose you read online that children ???rst 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 ???st count to 10 successfully is less than the general average of 32 months. Use a signi???cance level of 0.10.
Null hypothesis : The true population mean is 32
Alt hypothesis : The true population mean is not 32

n = 36
x = 32
mean = 30.69
sd = 4.31
standardError = sd/sqrt(n)
Z = (mean-x)/standardError
p = pnorm(Z,lower.tail = TRUE)
p
## [1] 0.0341013

A 3% chance of getting the observed mean if the proposed mean was true, MEANS (pun intended) we will reject the null hypothesis in favor of the alternative hypothesis.
#### (c) Interpret the p-value in context of the hypothesis test and the data.
If the null hypothesis were true, then we would have missed a 97% probability that it wasn’t.
####(d) Calculate a 90% con???dence interval for the average age at which gifted children ???rst count to 10 successfully

lowerTail = mean -   1.645 * standardError
upperTail = mean +   1.645 * standardError
lowerTail
## [1] 29.50834
upperTail
## [1] 31.87166

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

Yes, if the proposed mean was probable, it would not be higher than 31.87, but it is.

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 di???erent than the average IQ for the population at large, which is 100. Use a signi???cance level of 0.10.

null Hypothesis : ?? = 100 alt Hypothesis : ?? != 100

n = 36
mean = 118.2
sd = 6.5
x = 100
standardError = sd/sqrt(n)
Z = (mean-x)/standardError
Z
## [1] 16.8

Thats a big Z, probability around 0. Thats smaller than our significance level. I reject null hypothesis for the alternative hypothesis.
####(b) Calculate a 90% con???dence interval for the average IQ of mothers of gifted children.

lowerTail = mean - 1.645 * standardError
upperTail = mean + 1.645 * standardError
lowerTail
## [1] 116.4179
upperTail
## [1] 119.9821

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

Yes, the p-value is very small, and the proposed population mean is too far away from our confidence interval.

4.34 CLT. De???ne 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.

Sampling distribution of the mean, is when we repeatedly take samples of a population, and distribute the means we find from those samples into its own distribution.

4.40 CFLBs.

(a) 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

(b) Describe the distribution of the mean lifespan of 15 light bulbs.

Could be anything, sample size is not big enough to predict any outcome; central limit theorom applies, if anything it should be near normal distribution.
####(c) 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

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

x = 7000:11000
y = dnorm(x,mean=9000,sd=1000)
X = 7000:11000
Y = dnorm(X,mean = 9000,sd=258)
plot(x,y)

plot(X,Y)

####(e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution Nope, 30+ sample size needed for that.

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.

When you get your point estimates from a bigger sample, the variability, along with the P value gets smaller. So the P-value will decrease, pretty significantly from this mistake.