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?

Point estimate for average height is 171.1, and 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?

Point of estimate for the standard deviation for the heights is 9.4, and the IQR (inter quarilte range) is 14. (Q3 - Q1 = 177.8 - 163.8)

  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.

Those who are 180 cm and 155 cm tall are not considered unusually tall/short respectively because they are still within 2 standard deviations of the mean. Anything +/- 2 standard deviations from the mean is considered unusual.

  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.

Although the mean and standard deviations won’t be exactly the same, it is likely they will be similar due to the inherent nature of random sampling/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.

To meaure this, we use standard error to quantify the variability:

std_error <- 9.4/sqrt(507)
std_error
## [1] 0.4174687

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, because the confidence interval is built for the population mean, not the sample mean.

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

The distribution is right skewed, and so the confidence interval is not valid.

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

False, because if the random samples are at a different sample size than 436, then the intervals would be different.

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

True, because by definition the confidence interval covers the paramter values with 95% probability.

  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.

True

  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.

False, this is because standard error is inversely proportional to the square root of n–therefore, we would need to increase the sample size by 9X to decrease the margin of error.

  1. The margin of error is 4.4.

True

ZSe <- (89.11 - 80.31)/2
ZSe
## [1] 4.4

Gifted children, Part I. Researchers investigating characteristics of gifted children collected 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, because the sample size is greater than 30, the data follows a somewhat normal distribution, there are no outliers in the histogram, and the sample was collected from schools in a large city.

  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.

Null = 32 months; Alt = < 32 months; df = n-1 = 35; critical value = +/- 1.69; We reject the null if t < -1.69 or t > 1.69

The test statistics t = (30.69 - 32)/(4.31/sqrt(36)) = -1.824. We reject the NUll Hypothesis because -1.824 is less than -1.69. Therefore, at 0.10 level of significance, we conclude the average age which gifted children fist count to 10 successfully is less than 32 months.

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

2(1-P(t<(1.824)) = 2(1-0.9616) = 0.0767. Since 0.0767 is less than 0.10; therefore, we reject the null hypothesis.

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

The 90% confidence interval = mean ± margin error = 30.69 ± 1.69*(0.7183) = 30.69 ± 1.21 = (29.48, 31.9). Therefore, the 90% confidence we can say that the true average of age lies between 29.48 and 31.9 months.

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

Since the Null Hypothesis = 32 does not lie within the interval; therefore, the results from the hypothesis test and the confidence interval agree.

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 th eaverage 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 = 100 ; and Alternative Hypothesis DOES_NOT= 100. df = n-1 = 36-1 =35. Criticalvalue = +/- 1.69. . We reject the Null Hypothesis if: t < -1.69 or t > 1.69. . Determining test statistics; t = (118.2 - 100)/(6.5/sqrt(36)) = 16.8.

We reject the NUll Hypothesis because 16.8 is greater than 1.69. Therefore, at 0.10 level of significance, we can conclude that the average IQ of mothers of gifted children is different than the average IQ for the population at large, which is 100.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.3

90% confidence interval = mean +/- margin error = 118.2 +/- 1.69*(6.5/sqrt(36)) = 118.2 +/- 1.83 = (116.37, 120.03). Therefore, the 90% confidence indicates the true average of IQ is between 116.37 and 120.03.

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

Answer: Since the Null Hypothesis = 100 does not lie within the interval; therefore, the results from the hypothesis test and the confidence interval agree

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 sampling distribution of the mean is the probability distribution of the means for all possible random samples. As the sample size increases, the shape, center, and spread of the mean sampling distribution take the shape of a normal distribution.

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?

z = (10500 - 9000)/1000 = 1.5, P(Z>1.5) = 1 - P(Z<1.5) = 1 - 0.9332 = 0.0668. Therefore, it is 6.68% probability that a randomly chosen light bulb lasts more than 10,500 hours.

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

The lifespan mean follows a normal distribution. The sample mean is 9000, and the sample standard deviation is: 1000/sqrt(15) = 258.2

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

z = (10500 - 9000)/(1000/sqrt(15)) = 5.81, P(Z > 5.81) = 1 - P(Z<5.81) = 1 - 1.000 = 0. Therefore, there is no chance that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours.

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

sd <- 1000
mean <- 9000
se <- 1000/sqrt(15)

norm_sample <- seq(mean - (4 * sd), mean + (4 * sd), length=15)
random_sample<- seq(mean - (4 * se), mean + (4 * se), length=15)
norm <- dnorm(norm_sample,mean,sd)
random<- dnorm(random_sample,mean,se)

plot(norm_sample, norm, type="l",col="purple",
xlab="",ylab="",main="Population Distribution v Sample Distribution", ylim=c(0,0.002))
lines(random_sample, random, col="blue")

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

If the lifespans of the light bulbs had skewed distributions, we couldn’t estimate the probabilities because one of the assumptions for parts (a) and (c) cis that the data is normally distrubted..

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 normal probability, we observe that n is direcrtly proportional to the test statistic. So as the value of n increases, the corressponding test statistic will also. We also know that as the value of the test statistic increases, the P-value decreases. Therefore, the p-value increases when n increases from 50 to 500.