Graded: 4.4, 4.14, 4.24, 4.26, 4.34, 4.40, 4.48

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## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
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4.4

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.

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

Mean = 171.1 Median = 170.3

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

SD = 9.4cm IQR = 177.8-163.8 = 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.

A person 180cm tall would be in the top 25% of height, so yes they’re tall but not unusually tall. A person who is 155cm would be under the first quarter, and relatively close to the min. However, the 180cm student lands within one SD of the mean, while the 155cm student does not. I would consider the 155cm student unusually 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.

It’s very unlikely that the data will be the SD and mean would be the same due to the variability of the data of the sample chosen. It should be similar however.

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

SD = 9.4
n = 507
(SD_x = SD/sqrt(n))

## [1] 0.4174687

4.14

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 - the CI should be a rep of the population and not a sample
This confidence interval is not valid since the distribution of spending in the sample is right skewed. False - the data is only slightly skewed qnd the sample is so small this is invalid
95% of random samples have a sample mean between $80.31 and $89.11. False - samples can have different ranges therefor this is incorrect
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 CI, and what a CI measures
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 - the smaller the percentage of CI, the narrower the range
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 - it would need to be 9 times larger
The margin of error is 4.4. True

4.24

Researchers investigating characteristics of gifted children col- lected 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 dis- tribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.

Are conditions for inference satisfied? Yes; Randon sample, large sample, and no skew
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.

H_0 = 32 months H_a = < 32 months Sig level = 0.10

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

z = (30.69-32)/4.31 pnorm(-.3) = 0.38 p_val = 0.38

we fail to reject H_0 per the p_val of 0.38 > 0.1

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

MoE = 4.31/(sqrt(36))(-1.645.72) CI = (30.69-MoE,30.69+MoE) (29.51,31.87)

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

Yes, the CI of 90% is under 32 months, in what was our hypothesis from the beginning.

4.26

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.

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

mean = 118.2
n = 36
sd = 6.5
se = sd/sqrt(n)

z_score = (mean -100)/se
pnorm(z_score)

## [1] 1

We conclude to reject the null hypothesis; The avg IQ of the mother is different than that of the population

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

low = mean - 1.645*se
up = mean +1.645*se
low

## [1] 116.4179

up

## [1] 119.9821

(116.42,119.98)

Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, the results from the hypothesis test tell us that the avg IQ of the mother is different than that of the populaiton and also that the 90% CI is higher than the average IQ of the population at large

4.34

Define 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 - the distribution of n number of observations from a population. As the sample approaches the size of the population (i.e as n approaches the total number in the population), the sampling distribution becomes a normal distribution. The distribution takes a bell curve shape, and loses an skewness, modal, or even distribution it may have had before.

4.40

A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours.

What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

mean = 9000
sd = 1000
z_score = (10500 - 9000)/sd
p = 1-pnorm(z_score)
p

## [1] 0.0668072

Plot

normalPlot(bounds = c(z_score,10000000))

the probability is 6.68%

Describe the distribution of the mean lifespan of 15 light bulbs.

se_sample = 1000/sqrt(15)
se_sample

## [1] 258.1989

It would be normal

What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

z_score = (10500 - 9000)/258.2
prob = 1 - pnorm(z_score)
prob

## [1] 3.13392e-09

0% probability

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

normalPlot(mean = 9000, sd = 1000)

normalPlot(mean = 9000, sd = 258.2)

Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Since we assume normal distrubition in parts a and c, we would not be able to with a skewed distribution

4.48

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.

Since we’re increasing sample size to increase (ten fold) the spread of the distribution would be much more narrower, SD would also decrease with a larger n.Since H_0 is true, we would expect the value of p to decrease and thus strengthening our hypothesis.

DATA606 Homework 4