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?

    ## [1] "The point estimate for the average height of active individuals is  171.144"
    ## [1] "And the median is  170.3"

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

    ## [1] "The point estimate for the SD of heights of active individuals is  9.407"
    ## [1] "And the IQR is  14"

  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.

    We can get the Z-score for 180 cm and 155cm and if they are between -1 to 1; then we can say they are not unusal else they are more then 1SD apart from mean.

    ## [1] "Z-score for 180 cm tall is 0.941 ; which is between -1 and 1 and thus its NOT unusual"
    ## [1] "Z-score for 155 cm short is -1.716 ; which is NOT between -1 and 1 and thus its  unusual"

  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.

    The mean and SD from new sample could be similar but not necessarily the same; due to the samples are taken at random here.

  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.

    ## [1] "Standard Error is 0.42"

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.

    False- The point estimate is always in the confidence interval and not sample.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

    FALSE- This confidence level is not effected by this; the skew does not play an important role for the sample size.

  3. 95% of random samples have a sample mean between $80.31 and $89.11.

    False- different samples may have different range.

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

    True.

  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.

    True.

  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.

    False.

  7. The margin of error is 4.4.

    True.

    (89.11 - 80.31 ) / 2
    ## [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?

    Yes, Given the size n of 36 is large enough and the children were selected independent of each other at random. The plot does not show an skew and seems more of normal distribution.

  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.

    # Null hypothesis (H0): The average development for a child is {\sigma}=32 months.
# Alternate hypothesis (HA): The average development for a child is {\sigma}!=32 months.

# Calculate the Standard Error
Stand.Error <- sd(gifted$count)/sqrt(nrow(gifted ))

# Next we calcualte the Z-score
Zscore.32 <- (mean(gifted$count) - 32)/(Stand.Error)

# Calcualte the p-value as 
p.value.32 <- round((2*pnorm(Zscore.32, mean = 0, sd = 1)),2)

paste('The p value of',p.value.32,'is not equal to 0.10; hence we reject the Null hypothesis (H0)' )
    ## [1] "The p value of 0.07 is not equal to 0.10; hence we reject the Null hypothesis (H0)"

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

    We would reject the null hypothesis as the vlaue of p is low.

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

    lower.value <- round(mean(gifted$count) - 1.645 * Stand.Error,2)
upper.value <- round(mean(gifted$count) + 1.645 * Stand.Error,2)
paste(lower.value ,',',upper.value)
    ## [1] "29.51 , 31.88"

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

    Yes, they both reject the Null hypothesis.

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.

    # Null hypothesis (H0): Tthe average IQ of mothers of gifted children is different 
# than the average IQ for the population at large {\sigma}=100.

# Alternate hypothesis (HA): the average IQ of mothers of gifted children is equal 
# to the average IQ for the population at large {\sigma}!=100.

# Calculate the Standard Error
Stand.Error <- sd(gifted$motheriq)/sqrt(nrow(gifted ))

# Next we calcualte the Z-score
Zscore.100 <- (mean(gifted$motheriq) - 100)/(Stand.Error)

# Calcualte the p-value as 
p.value.100 <- round((2*(1-pnorm(Zscore.100, mean = 0, sd = 1))),2)

paste('The p value of',p.value.100,'is not equal to 0.10; hence we reject the Null hypothesis (H0)' )    
    ## [1] "The p value of 0 is not equal to 0.10; hence we reject the Null hypothesis (H0)"

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

    lower.value <- round(mean(gifted$motheriq) - 1.645 * Stand.Error,2)
upper.value <- round(mean(gifted$motheriq) + 1.645 * Stand.Error,2)
paste(lower.value ,',',upper.value)    
    ## [1] "116.38 , 119.95"

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

    Yes, they both 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.

Answer: The sampling distribution of the mean is distribution of the mean from many different samples from the population set. As the sample size increases the sampling distribution becomes closer to the Normal Distribution in its shape. The center does not change much and approaches the population mean. The spread narrows as sample size increases.

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?

    mean.hours <- 9000
sd.hours <- 1000

p.of.10500 <- round(pnorm(10500, mean=mean.hours , sd=sd.hours, lower.tail=F )*100, 2)

paste(p.of.10500, '% chance that a randomly chosen light bulb lasts more than 10,500 hours')
    ## [1] "6.68 % chance that a randomly chosen light bulb lasts more than 10,500 hours"

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

    # We can use the rnorm function here for the n=15 to generate the random deivates:
rnorm.15 <- rnorm(15 ,mean = mean.hours, sd = sd.hours)

# we can get the summary of this variable `rnorm.15` 
# and draw a historgram to see its distribution:

summary(rnorm.15)
    ##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    7475    7881    8263    8626    9372   10297
    hist(rnorm.15)

    # From above we can see the distribution is almost normal distributed.

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

    # Calculate the Standard Error
Stand.Error.15bulb <-  sd.hours/sqrt(15)

# Next we calcualte the Z-score
Zscore.15bulb <- (mean.hours - 15)/(Stand.Error.15bulb)

# Calcualte the p-value as 
p.value.15bulb <- round((2*(1-pnorm(Zscore.15bulb, mean = 0, sd = 1))),2)

paste('The p value of',p.value.15bulb,' is very small and probabality is thus 0%' )        
    ## [1] "The p value of 0  is very small and probabality is thus 0%"

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

    seq.x <- seq(0,12000,50)
dnorm.population <- dnorm(x = seq.x, mean = mean.hours, sd = sd.hours)
dnorm.sample <- dnorm(x = seq.x, mean = mean.hours, sd = Stand.Error.15bulb)
plot(x = seq.x, y = dnorm.sample, col = "blue", type = "l", xlab = 'Population vs Sampling', ylab = 'Density')
lines(x = seq.x, y = dnorm.population, col = "red")

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

    If the lifespans of light bulbs had a skewed distribution, then we cannot estiamte the probabilities.

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.

Answer: As the sample size increases; p value will decrease.