The mean is 171.1, and the median is 170.3
The point estimate for the standard deviation is 9.4, and the IQR is 7.5
Is a person who is 1m 80cm (180 cm) tall considered unusually tall?
No, the z-value of the observation at 180cm is approx. 1 standard deviation (0.95) above the mean and within the 95% confidence interval.
And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.
No, the z-value of the observation at 155cm is -1.71 standard deviations below the mean and within the 95% confidence interval.
#Observation of 180cm
sample_mean <- 171.1
observation <- 180
stand_error <- 9.4
(observation - sample_mean)/ stand_error ## [1] 0.9468085#Observation of 155cm
sample_mean <- 171.1
observation <- 155
stand_error <- 9.4
(observation - sample_mean)/ stand_error ## [1] -1.712766No. There are variations among different sample populations
To compute the Standard Error, take the Standard Deviation divided by the square root of the sample size which is 0.42.
stand_error / sqrt(507)## [1] 0.41746874.14 Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked o??? 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.
FALSE. We are 95% confident that the average spending for the population is within the range of ($80.31, $89.11). The sample average is a point estimate for the population.
FALSE. Given that the sample was chosen at random and that the sample size is greater than 30, we can apply the Central Limit Thereom to assume normality of the samples.
95% of random samples have a sample mean between $80.31 and $89.11. FALSE. The confidence interval is an estimate for the population not sample sizes.
We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
TRUE. The confidence interval shows that the true average spending for all Americans is between $80.31 and $89.11.
TRUE. At 90% confidence interval, the range would be narrower.
FALSE. In order to reduce the Standard Error by a third, the formula requires that the sample size be nine times the oringal sample since the formula is: sample mean * z-score * sqrt(sample size)
(89.11-80.31) / 2## [1] 4.44.24 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 distribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.43
Yes. Sample was chosen randomly. The sample is less than 10% of the population which means that it’s indepdendent, and the sample size is greater than 30 which can be approximated to be normal given the Central Limit Theorem.
Null Hypothesis: Students count to 10 when they’re 32 months old. Alternative Hypothesis: Students count to 10 when they’re less than 32 months old.
sample_mean2 <- 30.69
sample_std2 <- 4.31
sample_size2 <- 36
population_mean2 <- 32
stand_error2 <- (sample_std2) /sqrt(sample_size2)
stand_error2## [1] 0.7183333(sample_mean2 - population_mean2) / (sample_std2 / sqrt(sample_size2))## [1] -1.823666The z-score is -1.824 which corresponds to 0.9656. The p value is 0.0344 which is less than .10. Therefore, we can reject the null hypothesis.
1-0.9656## [1] 0.0344The p-value is the probability of observing data at least as favorable to the alternative hypothesis as our current data set. If the null hypothesis is true, the probability of observing a sample mean less than 32 months is only 3.4%
error <- 1.65 * stand_error2
left<- sample_mean2-error
right <- sample_mean2+error
left## [1] 29.50475right## [1] 31.87525A 90% Confidence Interval would be (29.50475, 31.87525)
Yes. The sample mean is 32 months, and although it’s at the border of the upper bound border for the 90% Confidence Interval, is not within the interval.. This along with the hypothesis test which rejected the null hypothesis both agree.
4.26 Gifted children, Part II. Exercise 4.24 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. ####Answers:
The Null Hypothesis is: IQ of Mothers of Gifted children = 100 The alternative hypothesis is: IQ of Mothers of Gifted children != 100
The z-score for the sample mean is 16.8. The p-score is the probability of getting that z-score < -16.8 is 0 which is less than .10. Thus, We can reject the null hypothesis
sample_mean3 <- 118.2
sample_std3 <- 6.5
sample_size3 <- 36
population_mean3 <- 100
(sample_mean3 - population_mean3) / ((sample_std3) / sqrt(sample_size3))## [1] 16.8stand_error3 <- (sample_std3) / sqrt(sample_size3)
error2 <- 1.65 * stand_error3
left2 <- sample_mean3 - error2
right2 <- sample_mean3 + error2
left2## [1] 116.4125right2## [1] 119.9875The 90% Confidence Interval for the sample population is 116.4125 to 119.9875.
4.34 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 sampling distribution represents the distribution of the point estimates (sample means) based on samples of a fixed size from a certain population. According to the Central Limit Theorem, the shape of the sample distribution becomes normal when the size of the individual samples are 30 or more. As the sample sizes get larger the variations (standard deviations) get smaller, but the means stay the same.
4.40 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. ####Answers:
z_score <- (10500 - 9000) / 1000
z_score## [1] 1.5probability = 1-0.9332
probability## [1] 0.0668ANSWER: Since the population is normally distributed, the distribution of the mean lifespan of the 15 light bulbs is also normally distributed.
ANSWER: The z-score of 15 light bulbs lasting 10,500 hours is 5.81 which translates into a probability of 1 under the normal curve. Thus, the probability of 15 bulbs lasting more than 10,500 hours is 0%
sample_size4 = 15
sample_mean4 = 10500
population_mean4 = 9000
population_sd = 1000
Z_score2 <- (sample_mean4-population_mean4) / (1000 / sqrt(15))
Z_score2## [1] 5.809475Sketch the two distributions (population and sampling) on the same scale.
Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
ANSWER: No, the estimates above are based on a the population and the samples being normally distributed under the Central Limit Theorem.
4.48 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.
When the sample size increases, the variability decreases. The absolute value of the Z score would also increase since the variability is lower. As the Z value increases, the p value decreases.