Suppose a random sample of 51 fatal crashes in 2015 in which driver had a positive blood alcohol concentration (BAC) results in a mean BAC of 0.158 gram per deciliter (g/dL) with a standard deviation 0.010 g/dL. Construct and interpret the 99% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. Use 4 non-zero decimal places in your calculations.
a. Find the t-alpha/2,df
b. Find and the margin of error (ME).
c. Find the Upper bound and the Lower bound. Interpret the confidence interval.
In: Statistics and Probability
In: Statistics and Probability
Correlation and Regression Analysis
Question
2
The marks in a Physics exam (P) and a Chemistry exam (C) were recorded for 15 students:
Physics |
25 |
46 |
63 |
45 |
78 |
18 |
84 |
48 |
73 |
50 |
61 |
89 |
38 |
36 |
30 |
Chemistry |
31 |
44 |
58 |
49 |
66 |
12 |
61 |
56 |
70 |
57 |
62 |
56 |
29 |
40 |
38 |
In: Statistics and Probability
A random sample of 100 adults were surveyed, and they were asked if the regularly watch NFL games. They were asked in their favorite team had ever won the Super Bowl.
Won Super Bowl | Did not win Super Bowl | Row total | |
Watch games | 24 | 56 | 80 |
Do not watch games | 11 | 9 | 20 |
Column total | 35 | 65 | 100 |
Find:
a) P(watch NFL games given their favorite team won the Super Bowl)
b) P(favorite team won the Super Bowl and watch NFL games)
c) P(favorite team won the Super Bowl or watch NFL games)
Please use and show excel formulas
In: Statistics and Probability
The TIV Telephone Company provides long-distance telephone service in an area. According to the company’s records, the average length of all long-distance phone calls placed through this company in 2015 was 12.44 minutes. The company’s management wants to check if the mean length of the current long- distance calls is different from 12.44 minutes. A sample of 150 such calls placed through the company produced a mean length of 13.71 minutes. The standard deviation of all such calls is 2.65 minutes. Using a 2% level of significance, can you conclude that the mean length of all current long-distance calls is different from 12.44 minutes? Hypothesis to be tested (in words): Hypothesis to be tested (using symbols): H0 : H1 : Is this problem a (circle one): left-tailed test, right tailed test, or a two-tailed test? a= ,x= ,sors= ,n= .
Test statistic =
The p-value =
Reject or fail to reject H0 (circle one) Rejection region (picture): Conclusion (in words):
In: Statistics and Probability
An investigator thinks that college students who pay for most of their own education study more than college students whose parents pay for most of their education. He gathers information on students regarding who is paying the greater portion of their college bills in order to put them into groups (parents pay (P) or student pays (S)) and also on the number of hours they report studying per week.
Choose the type of test you should use to test this hypothesis?
A) Z test
B) Single sample T test
C) Dependent T test
D) Independent T test
E) ANOVA
F) Pearson correlation T test
In: Statistics and Probability
Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained:; .SST= 10,780; SSTR= 4,520
a. Set up the ANOVA table for this problem (to 2 decimals, if necessary).
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value (to 4 decimals) |
Treatments | |||||
Error | |||||
Total |
b.Use to test for any significant difference in the means for the three assembly methods.
Calculate the value of the test statistic (to 2 decimals).
The -value is - Select your answer
-less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 12
What is your conclusion?
- Select your answer
-Conclude not all means of the three assembly methods are equalCannot reject the assumption that the means of all three assembly methods are equalItem 13
In: Statistics and Probability
To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material.
Manufacturer |
||||
1 | 2 | 3 | ||
25 | 33 | 20 | ||
31 | 31 | 19 | ||
29 | 36 | 23 | ||
27 | 32 | 22 |
a. Use these data to test whether the
population mean times for mixing a batch of material differ for the
three manufacturers. Use .
Compute the values below (to 2 decimals, if necessary).
Sum of Squares, Treatment | |
Sum of Squares, Error | |
Mean Squares, Treatment | |
Mean Squares, Error |
Calculate the value of the test statistic (to 2 decimals).
The -value is - Select your answer
-less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6
What is your conclusion?
- Select your answer
-Conclude the mean time needed to mix a batch of material is not the same for all manufacturersDo not reject the assumption that mean time needed to mix a batch of material is the same for all manufacturersItem 7
b. At the level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers and .
Calculate Fisher's LSD Value (to 2 decimals).
What is your conclusion about the mean time for manufacturer and the mean time for manufacturer ?
- Select your answer
-These manufacturers have different mean timesCannot conclude there is a difference in the mean time for these manufacturersItem 9
In: Statistics and Probability
Test the normality of the following phone call durations (in seconds), then replace each x value with log (x + 1) and test the normality of the transformed values. What does it concludes?
31.5, 75.9, 31.8, 87.4, 54.1, 72.2, 138.1, 47.9, 210.6, 127.7, 160.8, 51.9, 57.4, 130.3, 21.3, 403.4, 75.9, 93.7, 454.9, 55.1
In: Statistics and Probability
How productive are U.S. workers? One way to answer this question is to study annual profits per employee. A random sample of companies in computers (I), aerospace (II), heavy equipment (III), and broadcasting (IV) gave the following data regarding annual profits per employee (units in thousands of dollars).
I | II | III | IV |
27.3 | 13.9 | 22.7 | 17.8 |
23.9 | 9.4 | 20.5 | 16.6 |
14.4 | 11.5 | 7.1 | 14.2 |
8.1 | 8.7 | 12.9 | 15.4 |
11.2 | 6.8 | 7.7 | 10.2 |
19.2 | 9.5 |
Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the four types of companies? Use a 5% level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly three means are equal.Ho: μ1 = μ2 = μ3 = μ4; H1: All four means are different. Ho: μ1 = μ2 = μ3 = μ4; H1: Not all the means are equal.Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly two means are equal.
(b) Find SSTOT, SSBET, and
SSW and check that SSTOT =
SSBET + SSW. (Use 3 decimal places.)
SSTOT | = | |
SSBET | = | |
SSW | = |
Find d.f.BET, d.f.W,
MSBET, and MSW. (Use 3 decimal
places for MSBET, and
MSW.)
dfBET | = | |
dfW | = | |
MSBET | = | |
MSW | = |
Find the value of the sample F statistic. (Use 3 decimal
places.)
What are the degrees of freedom?
(numerator)
(denominator)
(c) Find the P-value of the sample test statistic.
P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
Since the P-value is greater than the level of significance at α = 0.05, we do not reject H0.Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H0. Since the P-value is greater than the level of significance at α = 0.05, we reject H0.Since the P-value is less than or equal to the level of significance at α = 0.05, we do not reject H0.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are all equal. At the 5% level of significance there is insufficient evidence to conclude that the means are all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.
(f) Make a summary table for your ANOVA test.
Source of Variation |
Sum of Squares |
Degrees of Freedom |
MS | F Ratio |
P Value | Test Decision |
Between groups | ---Select--- p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.001 < p-value < 0.010 p-value < 0.001 | ---Select--- Do not reject H0. Reject H0. | ||||
Within groups | ||||||
Total |
In: Statistics and Probability
#question1
library("olsrr")
##
## Attaching package: 'olsrr'
## The following object is masked from
'package:datasets':
##
## rivers
bass =
read.csv("bass.csv", header = T)
attach(bass)
bass.lm = lm(Avg_Mercury ~
Alkalinity + pH + Calcium
+ Chlorophyll, data = bass)
bass.all= ols_step_all_possible(bass.lm)
bass.all
## Index
N Predictors R-Square
Adj. R-Square
## 1 1
1 Alkalinity
0.4254905 0.4142256
## 2 2
1 pH
0.3310853 0.3179693
## 3 3
1 Calcium
0.2386129 0.2236838
## 4 4
1 Chlorophyll
0.2130176 0.1975865
## 6 5
2 Alkalinity
Calcium 0.4478582 0.4257726
## 7 6
2 Alkalinity
Chlorophyll 0.4436411 0.4213868
## 5 7
2 Alkalinity
pH 0.4292584 0.4064287
## 9 8
2 pH
Chlorophyll 0.3444788 0.3182580
## 8 9
2 pH
Calcium 0.3348995 0.3082955
## 10 10
2 Calcium
Chlorophyll 0.3009248 0.2729618
## 13 11 3 Alkalinity
Calcium Chlorophyll
0.4705171 0.4380997
## 11 12
3 Alkalinity
pH Calcium 0.4576077 0.4244001
## 12 13
3 Alkalinity
pH Chlorophyll
0.4436478 0.4095855
## 14 14
3 pH
Calcium Chlorophyll
0.3484270 0.3085347
## 15 15 4 Alkalinity pH Calcium Chlorophyll
0.4719492 0.4279450
## Mallow's Cp
## 1 3.223111
## 2 11.804576
## 3 20.210347
## 4 22.536973
## 6 3.189877
## 7 3.573211
## 5 4.880607
## 9 12.587099
## 8 13.457860
## 10 16.546176
## 13 3.130182
## 11 4.303642
## 12 5.572602
## 14 14.228213
## 15 5.000000
plot(bass.all)
detach(bass)
a. Give the ??2 and Adjusted??2 for the best models with one, two, three, and four predictors. Comment on these results (include the variables involved.)
b. Suppose that you want to predict the average mercury level of fish in a new lake with alkalinity 3.0, calcium 2.5, chlorophyll 2.5, and pH 6.0. The predicted value for the model including all four predictors is .545 (.0164, 1.073) [mean (PI).] The predicted value for the model including only alkalinity, calcium, and chlorophyll is .532 (.0133, 1.051). Have the predicted values and the prediction intervals changed considerably between the two models? Explain why or why not (based on the inspection of these results.)
c. Explain how your results of a) and b) agree.
In: Statistics and Probability
Consider the following table containing unemployment rates for a
10-year period.
Year | Unemployment Rate (%) |
---|---|
1 | 3.5 |
2 | 5.2 |
3 | 7.8 |
4 | 8.1 |
5 | 3.7 |
6 | 9.6 |
7 | 8.7 |
8 | 3.5 |
9 | 11.1 |
10 | 8.8 |
What is the coefficient of determination for the regression model? Round your answers to two decimal places.
In: Statistics and Probability
In: Statistics and Probability
what type of sampling procedure is most useful for participant observation studies A. probability B. Stratified C. snowball D. Cluster
In: Statistics and Probability
The University of Chicago claims to admit equal numbers of public and private high school graduates. You interview 500 undergraduates and discover 267 came from public schools. Assuming that the U of C is correct about its student body, what is the probability of seeing 267 or more students out of 500 interviewed come from public schools?
In: Statistics and Probability