Title: CUNY SPS MDS Data606_HW5"

Author: Charles Ugiagbe

Date: “10/6/2021”

Heights of adults. (7.7, p. 260) 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.

  1. What is the point estimate for the average height of active individuals? What about the median?
  2. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
  3. 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.
  4. 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.
  5. 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 \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Solution 1

summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1

  1. The point estimate for the mean height of active individuals is 171.1.

    The median is 170.3

  2. Point estimate for Standard deviation

sd(bdims$hgt)
## [1] 9.407205

Point estimate for Inter-quatile range

IQR <- IQR(bdims$hgt)
IQR
## [1] 14
  1. For a datapoint to be considered an outlier, it has to be lower than \(_{Q1}\) - 1.5IQR for low outlier or higher than \(_{Q3}\) + 1.5IQR for high outlier
Q1 = 163.8
Q3 = 177.8
lower_limit <- Q1 - 1.5*IQR
upper_limit <- Q3 + 1.5*IQR
limit <- range(lower_limit, upper_limit)
limit
## [1] 142.8 198.8

A 180cm tall person is not unusually tall because 180cm datapoint falls within the range and its not an outlier.

A 155cm tall person is not unusually tall because 155cm datapoint falls within the range; hence it’s not an outlier

  1. I will expect the mean and the standard deviation of the newly drawn sample to be close but not the same because they are drawn from the same population. This is because both the first and the second sample are ramdomly drawn from the population.

  2. We can quantify the variability of the estimate using:

\(SD_x = \frac{\sigma }{\sqrt{n}} = \frac{9.4}{\sqrt{507}} = 0.4175\)

Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked off 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.

  1. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.
  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed.
  3. 95% of random samples have a sample mean between $80.31 and $89.11.
  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
  5. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.
  6. 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.
  7. The margin of error is 4.4.

Solution 2:

  1. FALSE: The 95% confidence interval given as $80.31 and $89.11 is about the Population mean and not the sample mean.

  2. FALSE: The sample size of 436 is large enough to be model after a nearly normal distribution. Hence, we say the confidence interval is valid.

  3. FALSE: It’s erroneous to generalize that 95% ramdom sample have a sample mean between $80.31 and $89.11. A different sample size will give a different range of confident interval

  4. TRUE: The confidence interval is representing the Population of all American. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

  5. TRUE: The critical z value for a 99% confidence interval is 2.57 while that for a 95% confidence interval is 1.96 and for 90% confidence interval is 1.64. Hence, the The smaller the confidence interval the narrower the interval.

  6. FALSE: \(Margin of Error = Z^{*} * SE = Z^{*} * \frac{\sigma }{\sqrt{n}}\)

We need the sample size to be 9 times larger to decrease margin of error by 3

  1. TRUE \(Z^{*} * \frac{\sigma }{\sqrt{n}}\)

since sigma is unknown, we use:

Margin Error = (Upper Tail - Lower tail)/2

UpperTail <- 89.11
LowerTail <- 80.31
ME <- (UpperTail - LowerTail)/2
ME
## [1] 4.4

Gifted children, Part I. 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.

  1. Are conditions for inference satisfied?
  2. 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.
  3. Interpret the p-value in context of the hypothesis test and the data.
  4. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
  5. Do your results from the hypothesis test and the confidence interval agree? Explain.

Solution 3:

(a) Yes, the condition of inference are satisfied because:

  1. A randomize method was use to obtain the data from a large city (ii) The individual observation are independent of each other. (iii) The distribution is not overly skewed and sample size greater than 30. so, we say that the distribution is relatively normal.

(b) n = 36, mean = 30.69, sd = 4.31, df = (36 -1) =35

\(α=0.10\)

\(H0:μ=32\) , \(H1:μ<32\)

t.test(gifted$count, alternative = "less", mu = 32, conf.level = 0.90)
## 
##  One Sample t-test
## 
## data:  gifted$count
## t = -1.8154, df = 35, p-value = 0.03902
## alternative hypothesis: true mean is less than 32
## 90 percent confidence interval:
##     -Inf 31.6338
## sample estimates:
## mean of x 
##  30.69444

(c) The p-value of 0.039 < 0.10 which means that there is not enough evidence to support the null hypothesis. Hence we reject the null hypothesis and accept the alternative hypothesis

(d) SE = 4.31/SQRT(36) = 0.72

    30.87 +/- (z-score for 90% confidence interval * SE)
    
    30.69 +/- (1.645 * 0.72) = (29.51, 31.87)

the 90% confidence interval for the average age at which gifted children first count to 10 successfully is (29.51, 31.87)

(e) The result from the hypothesis testing of the p-value and confidence interval agreee because, they both result in a rejection of the mean. For the Confidence interval, the value of the mean fall outside the interval; hence the rejection.

Gifted children, Part II. Exercise above 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.

  1. 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.
  2. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
  3. Do your results from the hypothesis test and the confidence interval agree? Explain.

Solution 4:

(a) n = 36, mean = 118.2, sd = 6.5

\(\alpha = 0.10\) , \(DF = (36-1) = 35\)

\(H0:μ=100\) , \(H1:μ≠100\)

t.test(gifted$motheriq, alternative = "two.sided", mu = 100, conf.level = 0.90)
## 
##  One Sample t-test
## 
## data:  gifted$motheriq
## t = 16.756, df = 35, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 100
## 90 percent confidence interval:
##  116.3349 119.9984
## sample estimates:
## mean of x 
##  118.1667

The p-value of 2.2e-16 < 0.10 which means that there is not enough evidence to support the null hypothesis. Hence we cannot accept the null hypothesis.

(b)

    SE = 6.5/SQRT(36) = 1.08

 118.2 +/- (z-score for 90% confidence interval * SE)
 
 118.2 +/- (1.645 * 1.08) = (116.42, 119.98)

The population mean of 100 is outside the 90% confidence interval for the mean of Mother’s IQ with gifted children

(c) Yes, the both result Agree.

The P-value shows a big significant different between the from the confidence interval.hence, the sample mean is valid and we reject the null hypothesis. The confidence interval for the IQ of Mom’s with gifted kids is also way off the population mean of 100. Hence, with 90% confidence, we can also reject the null hypothesis.

CLT. 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.

Solution 5:

Sampling distribution of the mean is when we take random, independent samples of a constant sample size n. The graph showing the distribution of the values of the mean from all the samples taken is called the sampling distriubtion of the mean. The sampling distribution of the mean obeys the Central Limit Theorem in that it has a normal distirubtion (given sample size >= 30, and not overly skewed) and would tend towards the mean (sread becomes narrower) as sample size increases.

As the sample size increases -

  1. shape - closer to a normal distribution (normal curve)
  2. center - becomes taller (increase frequency of values that is close to the true population mean)
  3. spread - becomes narrower

CFLBs. 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.

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
  2. Describe the distribution of the mean lifespan of 15 light bulbs.
  3. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
  4. Sketch the two distributions (population and sampling) on the same scale.
  5. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Solution 6:

mean = 9000, sd = 1000, X = 10500

  1. Probability that a ramdomly chosen light bulb lasts more than 10,500 hours
P <- round(1 - pnorm(10500, mean = 9000, sd = 1000), 4)
P
## [1] 0.0668
  1. The mean lifespan of 15 light bulbs.
mean = 9000
sd = 1000
n = 15
SE = round(sd/sqrt(n), 2)
SE
## [1] 258.2

The population distribution is nearly normal \(N(9000, 1000^{2})\) and the SE for the sample is 258.2

  1. Probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
z <- 1 - pnorm(10500, mean = 9000, sd = SE)
z
## [1] 3.13392e-09

\(3.13392e-09 \simeq 0\) There is almost no chance. This is approximately Zero(0)

sd_pop <- 1000
sd_sample <- 258.2
x <- 5000:13000
dist_pop <- dnorm(x, mean = 9000, sd = sd_pop)
dist_sample <- dnorm(x, mean = 9000, sd = sd_sample)
plot(x, dist_pop, type = "l", main = "Distribution of compact fluorescent light bulbs",
     xlab = "Sample Vs Population", ylab = "Probabilities", col = "red", ylim = c(0, 0.0017))
lines(x, dist_sample, col = "blue")
legend(11300, 0.0012, legend = c("Population", "Sample"), fill = c("red", "blue"))

  1. The z values were based on the assumption of a normal distribution. Also, the sample size of 15 is small because we need sample size n >= 30 for central limit theorem to apply.

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.

Solution 7:

\(Z = \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{x- \mu}{\sigma}\sqrt{n}\)

As n increases so does t which casues the standard error and the p-value to get smaller.