assignment 1 utx

Sunny Patel IDS 371 Assignment 1 UIN: 653427443

1.

A) The stock ticket UTX is for United Technologies Corporation. According to Yahoo! Finance, United Technologies Corporation provides technology products and services to the building systems and aerospace industries worldwide. If you’ve ever seen “Otis” written on an elevator is because their Otis segment is responsible for designing, creating, and distributing freight elevators, escalators, and moving walkways. It is also one of the leading providers of heating, ventilation, A/C, and security. The company is now gearing towards improving its performance efficiently (eco-friendly).

B) Numerical Descriptive Statistics

Graphical Descriptive Statistics

0.04

0.03

0.02

0.01

0.00

-0.01

-0.02

-0.03

-0.04

R_UTX

Boxplot of R_UTX


0.03

0.02

0.01

0.00

-0.01

-0.02

-0.03

R_Q

QQ

Boxplot of R_ QQQ

0.02

0.01

0.00

-0.01

-0.02

R_DIA

Boxplot of R_DIA

0.03

0.02

0.01

0.00

-0.01

-0.02

-0.03

R_SPY

Boxplot of R_SPY


0.040.030.020.010.00-0.01-0.02-0.03

60

50

40

30

20

10

0

R_UTX

Frequency

Mean 0.0003192StDev 0.01232N 248

Histogram (with Normal Curve) of R_UTX

0.0270.0180.0090.000-0.009-0.018-0.027

40

30

20

10

0

R_QQQ

Frequency

Mean 0.0004603StDev 0.009536N 248

Histogram (with Normal Curve) of R_QQQ


0.02250.01500.00750.0000-0.0075-0.0150-0.0225

50

40

30

20

10

0

R_DIA

Frequency

Mean 0.0001736StDev 0.007284N 248

Histogram (with Normal Curve) of R_DIA

0.02250.01500.00750.0000-0.0075-0.0150-0.0225

50

40

30

20

10

0

R_SPY

Frequency

Mean 0.0003783StDev 0.007956N 248

Histogram (with Normal Curve) of R_SPY


C) 90% Confidence Intervals

0.02250.01500.00750.0000-0.0075-0.0150-0.0225

Median

Mean

0.00100.00050.0000-0.0005

1st Quartile -0.003896Median 0.0004013rd Quartile 0.004622Maximum 0.024621

-0.000456 0.001212

-0.000162 0.000768

0.007411 0.008595

A-Squared 1.47P-Value < 0.005

Mean 0.000378StDev 0.007956Variance 0.000063Skewness 0.022047Kurtosis 0.937943N 248

Minimum -0.025499

Anderson-Darling Normality Test

90% Confidence I nterval for Mean

90% Confidence I nterval for Median

90% Confidence I nterval for StDev90% Confidence I ntervals

Summary for R_SPY

R_SPY summary shows that it is symmetric with outliers along both sides of the graph.


0.0270.0180.0090.000-0.009-0.018-0.027

Median

Mean

0.00150.00100.00050.0000-0.0005


-0.000539 0.001460

-0.000602 0.001444

0.008882 0.010302

A-Squared 0.78P-Value 0.043


Minimum -0.026947


90% Confidence Interval for Mean

90% Confidence Interval for Median

90% Confidence Interval for StDev90% Confidence I ntervals

Summary for R_QQQ

R_QQQ shows the graph is symmetric with a few outliers also on each side of the map.

0.02250.01500.0075-0.0000-0.0075-0.0150-0.0225

Median

Mean

0.00100.00050.0000-0.0005-0.0010


-0.000590 0.000937

-0.000970 0.000606

0.006785 0.007869

A-Squared 1.30P-Value < 0.005


Minimum -0.021821


90% Confidence Interval for Mean

90% Confidence Interval for Median

90% Confidence Interval for StDev90% Confidence I ntervals

Summary for R_DIA

The R_DIA is skewed a bit to the left as its value is high and is less symmetric as the other.


D)

Using the hypothesis test (Ho: mu=0, Ha: mu not 0) we see that the returns for SPY, DIA, QQQ, & UTX are close to zero. It fails to reject Ho because the P-values (.455, .708, .448, .684) are higher than .10.

E&F) Confidence Intervals & P-Values given in each graph.


The two-sample T-test is used to compare the returns of six combinations of stock returns to 0 and then the test is used to compare the differences among the returns of the two chosen tickers. Once again in this situation, it fails to reject Ho because if you compare the P-values (.873,.887,.707, .949, .917, .765), they are all greater than .10. This means we can be sure that 90% of the time the difference among the given two ratios is zero.


2. Practice Exam Numbers 1-14 & 44

1. Other things being equal, the margin of error of a confidence interval increases as A) the sample size increases. B) the confidence level decreases. C) the population standard deviation increases. D) none of the above.

2. A study was conducted to compare the weights of sedentary workers with the weights of workers in physically demanding jobs. As part of the study, the weights of male accountants between the ages of 35 and 50 were recorded. Suppose a 99% confidence interval for the mean weight of accountants in pounds is (172.3, 176.5). If we had measured the weights of each of the accountants in kilograms (2.2 pounds = 1 kilogram), then the confidence interval for the mean weight of such accountants in kilograms would have been A) (174.5, 176.7). B) (78.32, 80.23). C) (379.06, 388.3). D) (170.1, 174.3).

3. Is the mean height for all adult American males between the ages of 18 and 21 now over 6 feet? If the population of all adult American males between the ages of 18 and 21 has mean height of P feet and standard deviation V feet, to answer this question one would test which of the following null and alternative hypotheses? A) H0: P = 6 vs. Ha: P > 6 B) H0: P = 6 vs. Ha: P < 6 C) H0: P = 6 vs. Ha: P z 6 D) H0: P = 6 vs. Ha: P = 6 ± x , assuming our sample size is n.

4. In a test of significance, assuming the null hypothesis is true, the probability that the test statistic will take a value at least as extreme as that actually observed is A) the P-value of the test. B) the level of significance of the test. C) the probability the null hypothesis is true. D) the probability the null hypothesis is false.

5. In a test of statistical hypotheses, the P-value tells us A) if the null hypothesis is true. B) if the alternative hypothesis is true. C) the largest level of significance at which the null hypothesis can be rejected. D) the smallest level of significance at which the null hypothesis can be rejected.


6. The nicotine content in cigarettes of a certain brand is normally distributed, with mean (in milligrams) P and standard deviation V = 0.1. The brand advertises that the mean nicotine content of its cigarettes is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a mean of x = 1.53. Is this evidence that the mean nicotine content is actually higher than advertised? To answer this, test the hypothesesH0: P = 1.5, Ha: P > 1.5 at the 5% significance level. You conclude A) that H0 should be rejected. B) that H0 should not be rejected. C) that Ha should be rejected. D) there is a 5% chance that the null hypothesis is true.

7. An agricultural researcher plants 25 plots with a new variety of corn that is drought resistant and hence potentially more profitable. The average yield for these plots is x = 150 bushels per acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown mean P and that a 95% confidence interval for P is found to be 150 ± 3.29. Which of the following is true? A) A test of the hypotheses H0: P = 150, Ha: P z 150 would be rejected at the 0.05 level. B) A test of the hypotheses H0: P = 150, Ha: P > 150 would be rejected at the 0.05 level. C) A test of the hypotheses H0: P = 160, Ha: P z 160 would be rejected at the 0.05 level. D) All the above.

8. The one sample t statistic from a sample of n = 19 observations for the two-sided test of

H0: P = 6, Ha: P z 6

has the value t = 1.93. Based on this information

A) we would reject the null hypothesis at D = 0.10. B) 0.025 < P-value < 0.05. C) we would reject the null hypothesis at D = 0.05. D) both b) and c) are correct.

9. The heights (in inches) of males in the U.S. are believed to be normally distributed with mean P. The average height of a random sample of 25 American adult males is found to be x = 69.72 inches and the standard deviation of the 25 heights is found to be s = 4.15.The standard error of x is A) 0.17. B) 0.69. C) 0.83. D) 2.04.


10. Scores on the SAT Mathematics test (SAT-M) are believed to be normally distributed with mean P. The scores of a random sample of three students who recently took the exam are 550, 620, and 480. A 95% confidence interval for P based on these data is A) 550.00 ± 173.88. B) 550.00 ± 142.00 C) 550.00 ± 128.58. D) 550.00 ± 105.01

An SRS of 100 postal employees found that the average time these employees had worked for the postal service was x = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately normal, with mean P. Are these data evidence that P has changed from the value of 7.5 years of 20 years ago? To determine this, we test the hypotheses

H0: P = 7.5, Ha: P z 7.5 using the one-sample t test.

11. The appropriate degrees of freedom for this test are A) 9. B) 10. C) 99. D) 100.

12. The P-value for the one-sample t test is A) larger than 0.10. B) between 0.10 and 0.05. C) between 0.05 and 0.01. D) below 0.01.

13. Suppose the mean and standard deviation obtained were based on a sample of 25 postal workers rather than 100. The P-value would be A) larger. B) smaller. C) unchanged because the difference between x and the hypothesized value P = 7.5 is unchanged.D) unchanged because the variability measured by the standard deviation stays the same.

14. A 95% confidence interval for the population mean time P that postal service employees have spent with the postal service is A) 7 ± 2. B) 7 ± 1.984. C) 7 ± 0.4. D) 7 ± 0.2.


Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. Net weights actually vary slightly from bag to bag and are normally distributed with mean P. A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised and so intends to test the hypotheses

H0: P = 14, Ha: P < 14.

To do this, he selects 16 bags of this brand at random and determines the net weight of each. He finds the sample mean to be x = 13.88 and the sample standard deviation to be s = 0.24.

44. Based on these data, we would conclude A) we would reject H0 at significance level 0.10 but not at 0.05. B) we would reject H0 at significance level 0.05 but not at 0.025. C) we would reject H0 at significance level 0.025 but not at 0.01. D) we would reject H0 at significance level 0.01.

Answers in order for Q1-14 & 441) C2) B3) A4) A5) D6) A7) C8) A9) C10) A11) C12) C13) A14) C44) B

assignment 1 utx

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