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?

Answer: The point estimate for the average height of active individuals is, mean = 171.1 and the median of active individuals is, median = 170.3.

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

Answer: The point estimate for the standard deviation of the heights of active individuals is, sd = 9.4 and the Inter Quartile Range (IQR)of the heights is, IQR = 177.8 - 163.8 = 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.

Answer: Data beyond 2 standard deviations away from the mean is considered unusual data. So, \(mean \pm SD = (171.1-(2*9.4), 171.1+(2*9.4)) = (171.1-18.8, 171.1+18.8) = (152.3,189.9)\). Therefore, 1m 80cm is within 2 standard deviation limits, so 1m 80cm is NOT considered as unusually tall. Same thing for 1m 55cm, it is within 2 standard deviation limits, so it is Not considered as unusually short.

  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.

Answer: The mean and standard deviation will not be EXACTLy the same but there is a high probability that the mean and standard deviation are very close to the ones given above. This is inherent 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.

Answer: We use the standard error measurement to quantify the variability. The \(SD_x = \frac{9.4}{\sqrt{507}}\) = 0.417.

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.

Answer: This statement is FALSE because 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.

Answer: This statement is FALSE because even though the distribution is right skewed but other conditions are met and thus confidence interval is perfectly valid here.

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

Answer: This statement is FALSE becasue if the random samples have a size different than 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.

Answer: This statement is TRUE because 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.

Answer: This statement is TRUE because 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.

Answer: This statement is FALSE. The Margin of Error = Critical value x Standard deviation and the \(standard error = \frac{\sigma}{\sqrt{n}}\). 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.

Answer: This statement is TRUE. We have sample mean + margin error = 89.11. Therefore, 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?

Answer: Yes. The conditions for inference was satisfied.

(1) The sample size is greater than 30.
(2) The histogram shape is near normal distribution.
(3) The histogram shows no strongly skewed or extreme outliers.
(4) Th sample was collected from schools located in a large city, which provides a population large enough for unbiased and independent.
  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.

Answer: Let the Null Hypothesis = 32 months, and Alternative Hypothesis < 32 months.

- Degrees of freedom = n-1= 36-1=35.
- \(Critical value = \pm 1.69.\)
- The Null Hypothesis can be rejected if the value of test statistics is either t < -1.69 or t > +1.69.
- The test statistics t = (30.69 - 32)/(4.31/sqrt(36)) = -1.824. The NUll Hypothesis is rejected since the -1.824 is less than -1.69. Therefore, at 0.10 level of significance, we can conclude that the average age which gifted children fist count to 10 successfully is less than the general average of 32 months.
  1. Interpret the p-value in context of the hypothesis test and the data.

Answer: The p-value at two tails of the test = 2(1-P(t<(1.824)) = 2(1-0.9616) = 0.0767. Since 0.0767 is less thatn 0.10; therefore, the Null Hypothesis is rejected.

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

Answer: The 90% confidence interval = mean \(\pm\) margin error = 30.69 \(\pm\) 1.69*(0.7183) = 30.69 \(\pm\) 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.

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

Answer: 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. 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.

Answer: Let the Null Hypothesis = 100 , and Alternative Hypothesis \(\neq\) 100.

- Degrees of freedom = n-1= 36-1=35.
- \(Critical value = \pm 1.69.\)
- The Null Hypothesis can be rejected if the value of test statistics is either t < -1.69 or t > +1.69.
- The test statistics t = (118.2 - 100)/(6.5/sqrt(36)) = 16.8. The NUll Hypothesis is rejected since the 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.

Answer: The 90% confidence interval = mean \(\pm\) margin error = 118.2 \(\pm\) 1.69*(6.5/sqrt(36)) = 118.2 \(\pm\) 1.83 = (116.37, 120.03). Therefore, the 90% confidence we can say that the true average of age lies between 116.37 and 120.03 IQ.

  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.

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 mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean μ, then the mean of the sampling distribution of the mean is also μ. The symbol \({ \mu }_{ M }\) 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: \({ \mu }_{ M } = μ\). 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 finaly 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?

Answer: 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.

Answer: The mean of lifespan follows normal distribution. Th mean of sampling means is 9000 and the standard deviation of sample mean = 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?

Answer: z = (10500 - 9000)/(1000/sqrt(15)) = 5.81, P(Z > 5.81) = 1 - P(Z<5.81) = 1 - 1.000 = 0. Therefore, it is 0% probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours.

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

Answer:

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

normsample <- seq(mean - (4 * sd), mean + (4 * sd), length=15)
randomsample<- seq(mean - (4 * se), mean + (4 * se), length=15)
hnorm <- dnorm(normsample,mean,sd)
hrandom<- dnorm(randomsample,mean,se)

plot(normsample, hnorm, type="l",col="blue",
xlab="",ylab="",main="Distribution of Population vs Sampling", ylim=c(0,0.002))
lines(randomsample, hrandom, col="red")

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

Answer: If the lifespans of light bulbs had a skewed distribution, we cannot estimate the probabilities because one of the assumptions in order to perform parts (a) and (c) calculations is that it must be a normal distribution.

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: From the z-test statistic formula, \(z = (x - μ)\ \frac{\sigma}{\sqrt{n}}\), 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.