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?
head(bdims$hgt)
## [1] 174.0 175.3 193.5 186.5 187.2 181.5
mean_h <- mean(bdims$hgt)
mean_h
## [1] 171.1438
median_h <- median((bdims$hgt))
median_h
## [1] 170.3

The point estimate for the average height is 171.1. The point estimate for the median is 170.3.

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

IQR = 14

sd_h <- sd(bdims$hgt)
round(sd_h, 1)
## [1] 9.4
IQR(bdims$hgt, na.rm = T)
## [1] 14

The point estimate for the standard deviation of the heights is 9.4. The point estimate for the IQR is 14.

  1. 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.
pos <- mean_h  + 2*sd_h 

neg <- mean_h  - 2*sd_h 
pos
## [1] 189.9582
neg
## [1] 152.3294
pos2 <- mean_h + 1*sd_h  

neg2 <- mean_h - 1*sd_h  


pos2
## [1] 180.551
neg2
## [1] 161.7366

An individual 180cm high is not unusually considered tall. This is because he is just 0.95 above the standard deviation. An individual 155cm high is not considered unusually short. This is because he is just 1.71 below the standard deviation. Both cases/instances are within two standard deviation from the mean.

  1. 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.
hist(bdims$hgt, probability = TRUE)
x <- 140:200
y <- dnorm(x = x, mean = mean_h, sd = sd_h)
lines(x = x, y = y, col = "blue")

The mean and standard deviation will not be precisely the same, yet there is a high probability that the mean and standard deviation are close enough to the values given above. This is due to to random sampling and random distributions.

  1. 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.
(x <- sd_h/sqrt(nrow(bdims)))
## [1] 0.4177887

The standard we would apply for the variability is the standard error.

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.

library(openintro)

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

The statement is false since confidence interval is always constructed for the population mean, and not for the sample mean.

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

The statement is false since even though the distribution is right skewed, still other conditions are met and therefore confidence interval is perfectly valid here.

  1. 95% of random samples have a sample mean between $80.31 and $89.11.

The statement is false since when the random samples have a size different from 436 randomly sampled American adults, then the intervals will be different.

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

The statement is true for this is what the confident interval should be interpreted.

  1. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

The statement is true for the critical value is less at 90% confidence interval compared to the 95% confidence interval.

  1. 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.

The statement is false. The Margin of Error = Critical value x Standard deviation and the standard error=σn√. As you can see from the standard error formula which standard error is inversely proportional to the square root of n. Therefore, in order to decrease the margin of error of a 95% confidence level to a third, we have to increase the sample size by 9 times, not 3 times larger only.

  1. The margin of error is 4.4.

This statement is true. We have sample mean + margin error = 89.11. Hence, the margin error = 89.11 - 84.71 = 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?

The conditions for inference are satisfied since a random sample was taken,independence is satisfied, the sample size is over 30, and the distribution is not extremely skewed.

  1. 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.
mean <- 30.69
n <- 36
sd <- 4.31
x <- 32
se <- sd / sqrt(n)
Z_score <- (mean - x) / se
(p <- pnorm(Z_score, mean = 0, sd = 1))
## [1] 0.0341013
round(p, 3)
## [1] 0.034

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

The p-value in context of the hypothesis and the data is since the p-value is 0.03 and is lower than the significance level of 0.10, we reject the null hypothesis that the average age at which gifted children first count to 10 is the same as the general population and accept the alternative hypothesis which is the average age at which gifted children count to 10 is less than the general population.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
lower <- mean - 1.645 * se
upper <- mean + 1.645 * se
c(lower, upper)
## [1] 29.50834 31.87166

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

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

The results from the hypothesis test and the confidence interval do agree because the hypothesis test tells us to reject the null hypothesis that the average age is 32 months and the 90% confidence interval does not include 32 months.

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. Performahypothesistesttoevaluateifthesedataprovideconvincingevidencethattheaverage 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.
mean_2 <- 118.2
n_2 <- 36
sd_2 <- 6.5
x_2 <- 100
se_2 <- sd_2 / sqrt(n_2)
Z_score_2 <- (mean_2 - x_2) / se_2
p_2 <- (1 - pnorm(Z_score_2, mean = 0, sd = 1)) * 2
p_2
## [1] 0

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
lower_2 <- mean_2 - 1.645 * se_2
upper_2 <- mean_2 + 1.645 * se_2
c(lower_2, upper_2)
## [1] 116.4179 119.9821

The 90% confidence interval for the average IQ of mothers of gifted children is (116.42, 119.98).

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

The results from the hypothesis test and the confidence interval do agree because the hypothesis test tells us to reject the null hypothesis that the average IQ of mother is 100 and the 90% confidence interval doesn not include 100.

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.

The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Hence, if a population has a mean μ, then the mean of the sampling distribution of the mean is also \({μ}\). The symbol \({μ_M}\) is used to refer to the mean of the sampling distribution of the mean. Therefore, the formula for the mean of the sampling distribution of the mean can be written as: \({μ_M \ = \ μ}\). The shape of the sampling distribution will tend to be inverted V shaped as sample size increases, the spread will tend to be lower with increase in sample size, and finally the center will tend to coincide with the true mean of the population, as the 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?
1 - pnorm(q = 10500, mean = 9000, sd = 1000)
## [1] 0.0668072

The probability that a randomly chosen light bulb lasts more than 10,500 hours is 0.067 or 6.7%.

  1. Describe the distribution of the mean lifespan of 15 light bulbs.
1000 / sqrt(15)
## [1] 258.1989

he distribution of the mean lifespan of 15 light bulbs is nearly normal with a mean of 9000 and a standard deviation of 258.20.

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
round(1 - pnorm(q = 10500, mean = 9000, sd = 258.20), 4)
## [1] 0

The probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours is basically 0%.

  1. Sketch the two distributions (population and sampling) on the same scale.
x <- 4000:14000
population <- dnorm(x, mean = 9000, sd = 1000)
sampling <- dnorm(x, mean = 9000, sd = 258.20)
plot(x, population, type = "l", col = "red")
lines(x, sampling, col = "blue")

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

You could not estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution because you would need a normal distribution and for part (c) the sanple size is too small.

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.

From the z-test statistic formula, z=(x−μ) σn√, we can clearly observe that n is direcrtly proportional to test statistic. So as the value of n increases the corresponding test statistic will also increase. As we know that as the test statistic’s numerical value increases P-value decreases. Therefore, the p-value increases when n change from 50 to 500.