Download Hypothesis Testing and Confidence Intervals and more Exams Nursing in PDF only on Docsity!
STAT 200: Introduction to
Statistics Homework
Solutions
- (3 points): Stephen Stigler determined in 1977 that the speed of light is 299,710.5 km/sec. In 1882, Albert Michelson had collected measurements on the speed of light ("Student t-distribution," 2013). Is there evidence to show that Michelson’s data is different from Stigler’s value of the speed of light? a.) State the random variable For this problem, the random variable will be: x = speed of light measured by Albert Michelson
b.) State the population parameter The population parameter will be: μ = mean speed of light measured by Albert Michelson
c.) State the hypotheses The hypotheses for this experiment are given by: 𝑯𝟎: 𝝁 = 𝟐𝟗𝟗, 𝟕𝟏𝟎. 𝟓 𝒌𝒎/𝒔 𝑯𝑨: 𝝁 ≠ 𝟐𝟗𝟗, 𝟕𝟏𝟎. 𝟓 𝒌𝒎/𝒔
- (3 points): According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? a.) State the type I error in this case, consequences of this error type for this situation, and the appropriate alpha level to use. In this situation, the Type I error is saying that the proportion of complaints from identity theft in Alaska is less than 23%, when it is 23%. One consequence of this error is that the Federal Trade Commission (FTC) would think that identity theft isn’t as big as a problem when it is.
Thus, the FTC may not put as much effort into stopping or investigating identity theft in Alaska as it should.
b.) State the type II error in this case, consequences of this error type for this situation, and the appropriate alpha level to use.
Type II error: saying that the proportion of complaints from identity theft in Alaska is 23%, when it is less than 23%. One consequence of this error is that the Federal Trade Commission would put more effort into Alaska then it needs to.
Thus, resources that could be used other places will be wasted in Alaska.
The best alpha level in this case would be 1%, since a type I error looks to have worse consequences than a type II error.
n = 1432^4
𝑝 = 𝑥^ = 321 = 0. 𝑛 1432
The test statistic is given by:
𝑝 − 𝑝 0.2242 − 0. 𝑧 = = = −0. 𝑝𝑞 √ (^) 𝑛
√0.23(1^ −^ 0.23)
The p-value associated with this problem (going back to homework 4 for how to compute the p-value from a z-statistic) is given by:
= NORM.S.DIST(z,cumulative)
= NORM.S.DIST (-0.522, TRUE) = 0.
v.) Conclusion^5
Since the p-value is greater than the level of significance (i.e. [p-value =
0.2998] > [α = 0.05]), we fail to reject 𝑯𝟎.
vi.) Interpretation (do not skip this part! This is the “so what” of the entire hypothesis test).
There is not enough evidence to show that the proportion of complaints due to
identity theft in Alaska is less than 23%.
- (3 points): In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are diagnosed with ASD ("CDC features -," 2013). Is there sufficient data to show that the incident of ASD is more in Arizona than nationally? Why or why not? Test at the 1% level.
We should start by writing down what we know (which is always a great place to start): x = 507 n = 32, p = 1/88 = 0.0114 (or 1.14%) α = 0.
To fully address this problem, we should follow the six step process presented in the textbook. i.) State the random variable and the parameter in words. The random variable is given by: x = number of children in Arizona in 2008 that were diagnosed with Autism Spectrum Disorder (ASD) The parameter of interest is given by: p = proportion of children in Arizona in 2008 that were diagnosed with Autism Spectrum Disorder (ASD)
ii.) State the null and alternative hypotheses and the level of significance The hypotheses for this experiment are given by: 𝐻 : 𝑝 = 1 = 0. (^0 ) 𝐻 : 𝑝 > 1 = 0. 𝐴 (^88) The level of significance is α = 0.01.
iv.) Find the sample statistic, test statistic, and p-^7 value The sample proportion is given by: x = 507 n = 32, 𝑝 = 𝑥^ = 507 = 0. 𝑛 32,
The test statistic is given by: 𝑝 − 𝑝 0.0156 −^ 0. 𝑧 = = = 7. 𝑝𝑞 0.0114(1 − 0.0114) √ (^) 𝑛 √ 32601 The p-value associated with this problem (going back to homework 4 for how to compute the p-value from a z-statistic) is given by: =1 - NORM.S.DIST(z,cumulative) =1 - NORM.S.DIST (7.134, TRUE) = 4.866 * 10-^13
v.) Conclusion
Since the p-value is less than the level of significance (i.e. [p-value = 4.866 *
10 -^13 ] < [α = 0.01]), we reject 𝑯𝟎.
vi.) Interpretation (do not skip this part! This is the “so what” of the entire hypothesis test).
There is enough evidence to show that the proportion of Arizona children in
2008 with ASD is more than the national proportion.
- (3 points): The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in Table 1 ("SOCR data 2008," 2013). Countries that are considered high-income have a mean economic dynamism of 60.29. Does the data show that the mean economic dynamism of middle-income countries is less than the mean for high income countries? Why or why not? Test at the 5% level.
25.8057 37.4511 51.9150 43.6952 47.8506 43.7178 58. 41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43. 49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43. 50.9866 59.1724 39.6282 33.6074 21.
Table 1: Economic Dynamism of Middle Income Countries
i.) State the random variable and the parameter in words.
x = economic dynamism for a middle-income country μ = mean economic dynamism for middle-income countries
ii.) State the null and alternative hypotheses and the level of significance 𝐻 0 : 𝜇 = $60. 𝐻𝐴: 𝜇 < $60. 𝛼 = 0.
√^ √
Normal Probability Plot for Economic Dynamism
2
1
0
- 00
iv.) Find the sample statistic, test statistic, and p- value Sample mean and standard deviation: 𝑥 = $43. 𝑠 = $9. n = 26
Test Statistic: 𝑥 − 𝜇 𝑡 = (^) 𝑠
= 9.07 =^ −9.
⁄ 𝑛 ⁄^26
p-value: To get the p-value from excel, we use the t.dist function:
Syntax: T.DIST(x,deg_freedom, cumulative)
The T.DIST function syntax has the following arguments: X Required. The numeric value at which to evaluate the distribution Deg_freedom Required. An integer indicating the number of degrees of freedom. Cumulative Required. A logical value that determines the form of the function. If
0000 10.00 00 20.00 00 30.00 00 40.00 00 50.00 00 60.00 00 70.
Normal Probabiltiy Plot for Sway
2
1
0
- (3 points): Maintaining your balance may get harder as you grow older. A study was conducted to see how steady the elderly is on their feet. They had the subjects stand on a force platform and have them react to a noise. The force platform then measured how much they swayed forward and backward, and the data is in table #7.3.10 ("Maintaining balance while," 2013). Does the data show that the elderly sway more than the mean forward sway of younger people, which is 18.125 mm? Why or why not? Test at the 1% level.
19 30 20 19 29 25 21 24 50
Table 2: Forward/backward Sway (in mm) of Elderly Subjects
i.) State the random variable and the parameter in words. x = forward and backward sway of an elderly person μ = mean forward and backward sway of an elderly person
ii.) State the null and alternative hypotheses and the level of significance 𝐻 0 : 𝜇 = 18.125 𝑚𝑚 𝐻𝐴: 𝜇 > 18.125 𝑚𝑚 𝛼 = 0.
iii.) State and check the assumptions for a hypothesis test a) A simple random sample of the forward and backward sway of 9 elderly people was taken. The problem doesn’t mention how the sample was taken. So this requirement may not have been met. b) The population of the forward and backward sway of all elderly people is normally distributed. The histogram does not look bell shaped, there is one outlier, and the normal probability plot does not appear linear. Thus, this assumption may not be met.
0 10 20 30 40 50 60
0.52 p 0.60, where p is the proportion of Americans who believe it is the government’s responsibility for health care. Give the statistical interpretation.
The proportion of Americans who believe it is the government’s
responsibility for health care is between 52% and 60%.
- (3 points): In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%.
This is a confidence interval about a proportion. Thus, we will use the standard normal distribution.
i.) State the random variable and the parameter in words. x = number of children in Arizona in 2008 that were diagnosed with Autism Spectrum Disorder (ASD) p = proportion of children in Arizona in 2008 that were diagnosed with Autism Spectrum Disorder (ASD)
ii.) State and check the assumptions^17 a. A simple random sample of the 32,601 diagnoses of children was taken in 2008. The study was conducted by the CDC, so this assumption is probably true. b. There are 32,601 diagnoses in this sample. The diagnoses of one Arizona child doesn’t affect the opinion of the next one. There are only two outcomes, either the Arizona child has ASD or they do not. The chance that one Arizona child has ASD does not change. Thus, the conditions for the binomial distribution are satisfied c. In this case, 𝑝 = 𝑥^ = 507 = 0.0156^ and^ n^ = 32601. 𝑛 32, Thus, n 𝑝 = 32601 * 507 32,
= 507 ≥ 5 and n 𝑞= 32601 * (32,601−507) = 32094 ≥ 5. 32,
Thus, the sampling distribution for 𝑝 is a normal distribution.
iv.) Find the sample statistic and confidence
interval The sample proportion is given by: x = 507 n = 32, 𝑝 = 𝑥^ = 507 = 0. 𝑛 32,
Confidence Interval: First, we need to determine the value for 𝑧𝑐, the critical value where C = 1 – α If we use Table A.1 in the back of the Kozak textbook, we find this value is 2.575.
Table A.1: Normal Critical Values for Confidence Levels
Confidence Level, C Critical Value, zc 99% 2. 98% 2. 95% 1. 90% 1. 80% 1.
You might actually want to know from where this value came, so here is how you can find it in Excel: Since we are looking at the 99% confidence interval, we have an area of 1 – 0.99 = 0.01 outside of our confidence interval; however, half is on both sides of the interval. Thus, it goes from 0.005 to 0.995.
The last step is to put this into the confidence interval equation:^19 𝑝 − 𝐸 < 𝑝 < 𝑝 + 𝐸 0.01555 − 0.0018 < 𝑝 < 0.01555 + 0. 0.01379 < 𝑝 < 0.
iv). Statistical Interpretation: There is a 99% chance that the interval 𝟎. 𝟎𝟏𝟑𝟕𝟗 < 𝒑 < 𝟎. 𝟎𝟏𝟕𝟑𝟐 contains the true proportion of children in Arizona in 2008 that were diagnosed with Autism Spectrum Disorder (ASD).
v.) Real World Interpretation: The proportion of children in Arizona in
2008 that were diagnosed with Autism Spectrum Disorder (ASD) is
between 0.01379 and 0.017321.
- (3 points): The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in Table 1 ("SOCR data 2008," 2013). NOTE: this is the same data set from question 5. Compute a 95% confidence interval for the mean economic dynamism of middle-income countries.
25.8057 37.4511 51.9150 43.6952 47.8506 43.7178 58. 41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43. 49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43. 50.9866 59.1724 39.6282 33.6074 21.
Table 1: Economic Dynamism of Middle Income Countries
This is a confidence interval about the mean, when the population mean is NOT known. Thus, we will use Student’s t distribution.
i.) State the random variable and the parameter in words. x = economic dynamism for a middle-income country p = mean economic dynamism for middle-income countries
ii.) State and check the assumptions
a. A simple random sample of economic dynamism for 26 middle-income countries was taken.^20 The problem doesn’t mention how the sample was taken. Thus, this assumption may not have been met. b. Recall from question 5: The population of the economic dynamism for all middle-income countries is normally distributed or the sample size is 30 or more. The sample size is 26. The histogram looks somewhat bell shaped, there is one outlier (but it is not far outside 1.5*IQR), and the normal probability plot does appear linear. Thus, this assumption is probably met (nothing is ever “perfect” in real life).
iv.) Find the sample statistic and confidence interval Also from question 5: Sample mean and standard deviation: 𝑥 = $43. 𝑠 = $9. n = 26
