Data collected by a price reporting agency from more than 90,000 gasoline and convenience stores throughout the U.S. showed that the average price for a gallon of unleaded gasoline was $3.28. The following data show the price per gallon ($) for a sample of 20 gasoline and convenience stores located in San Francisco.
| 3.59 | 3.59 | 4.59 | 3.76 | 3.75 | 3.71 | 3.65 | 3.80 | 3.55 | 3.36 |
| 3.57 | 3.79 | 3.55 | 4.19 | 3.95 | 3.46 | 3.83 | 3.93 | 3.41 | 3.37 |
a) Use the sample data to estimate the mean price for a gallon of unleaded gasoline in San Francisco.
b) Compute the sample standard deviation. (Round your answer to the nearest cent.)
c) Compare the mean price per gallon for the sample data to the national average price. What conclusions can you draw about the cost of living in San Francisco?
The average price for a gallon of unleaded gasoline in San Francisco is ---Select--- higher than lower than same as the national average. This indicates that the cost of living in San Francisco is ---Select--- higher than lower than same as it would be for cities that have an average price close to the national average.
In: Math
Find the indicated probabilities using a standard normal
distribution
a. P(Z < 1.85)
b. P(Z < -1.54 or Z > 1.54)
In: Math
The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a 95% confidence interval estimate of the difference between type 1 and type 2 yields. What does the confidence interval suggest about farmer Joe's claim that type 1 seed is better than type 2 seed? Type 1 1988 2014 2190 2478 2208 1983 2258 1504 Type 2 2012 1919 2099 2437 2167 1911 2179 1444 In this example, mu Subscript d is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the type 1 seed yield minus the type 2 seed yield. The 95% confidence interval is 24.66 less thanmu Subscript dless than89.09 . (Round to two decimal places as needed.) What does the confidence interval suggest about farmer Joe's claim that type 1 seed is better than type 2 seed? A. Because the confidence interval only includes positive values and does not include zero, there is sufficient evidence to support farmer Joe's claim. This is the correct answer. B. Because the confidence interval includes zero, there is sufficient evidence to support farmer Joe's claim. C. Because the confidence interval only includes positive values and does not include zero, there is not sufficient evidence to support farmer Joe's claim. Your answer is not correct. D. Because the confidence interval includes zero, there is not sufficient evidence to support farmer Joe's claim.
In: Math
Some researchers tested whether arthritis in dogs could be improved by supplementation with antioxidants and/or an aminosugar mixture (containing glucosamine and chondroitin). They gave combinations of these supplements (each a factor with two levels: treatment and control) to equal numbers of test subjects in a balanced factorial design. They tested the effects of these supplements on levels of inflammation using a factorial ANOVA. The ANOVA table from their output is copied below.
> model<-aov(inflammation~as.factor(antioxidant)*as.factor(aminosugar))
> anova(model)
|
Df |
SS |
MS |
F |
P |
|
|
antioxidant |
1 |
385 |
385 |
17.5 |
0.0007 |
|
aminosugar |
1 |
0.7 |
0.7 |
0.032 |
0.8581 |
|
antioxidant:aminosugar |
1 |
1.3 |
1.3 |
0.059 |
0.7863 |
|
Residual |
16 |
352 |
22 |
a) How many hypotheses did they test with this model?
b) How many test subjects (i.e. replicates) did the researchers have?
c) Did the order in which antioxidant and aminosugar effects entered this model affect their significance? Why?
d) Name a measure of model fit that can be used to compare the relative fit of different models, while taking into account the number of parameters in each?
e) The researchers simplified their model by removing the interaction term and the main effect of aminosugar. Fill in the table below with the values of their new model.
> model2<-aov(inflammation~as.factor(antioxidant))
> ANOVA(model)
|
Df |
SS |
MS |
F |
|
|
antioxidant |
||||
|
Residual |
f) Did the significance of antioxidant change by removing the other terms, and if so, did it become more or less significant?
g) By how much did the overall model R2 change (explain whether it increased, decreased or no change, as well as the amount)? (show your working)
In: Math
Boneitis is in the news again: several cases have been reported in Miami, and more are soon to follow. In order to receive aid from the state government, the city of Miami has to provide a small amount of evidence that the prevalence of the disease (i.e., the proportion of Miami residents who have boneitis) is greater than 20%. City ocials took a random sample of 200 residents and found that 33.0% had been diagnosed with boneitis.
(a) (4 points) Perform a z-test at the 0.01 significance level to determine if this sample provides evidence that the prevalence of boneitis in Miami is greater than 20%. Be sure to state the null and alternative hypotheses, any calculator functions you use (including inputs and relevant outputs), and clearly state your conclusion in context.
(b) (3 points) Construct a 95% plus-two confidence interval for pˆ, the sample proportion of Miami residents with boneitis. Show your work, including any calculator functions you use, and be sure to round interval bounds properly.
In: Math
Compounds in the saliva, hair and urine of pets may potentially trigger asthma in some patients. To test this hypothesis, some researchers conducted a study. As test subjects, they used 8 families, 4 of which had a pet in their home and 4 did not. They measured respiratory flow of each family member, as this is known to decrease in patients with asthma. Recent research has shown body weight to correlate with asthma in children, and sex, age, and height are also known to affect respiratory flow measures. Therefore, the researchers measured and included these covariates in their model to control for any confounding effects that they may have had before testing their treatment of interest (whether the home had pets). The analysis conducted by the researchers is presented below. They concluded that pets in the family do not influence respiratory flow.
> attach(pets.data)
> pets.data
resp.flow body.weight treatment sex age height family
1 266 41 pet F 13 130 Smith
2 400 95 pet M 37 190 Smith
3 369 81 pet F 36 180 Smith
4 365 78 pet M 42 185 Wilson
5 351 75 pet M 35 192 Wilson
6 334 62 pet F 41 170 Taylor
...
13 391 78 no pet M 41 180 Singh
14 340 62 no pet F 27 177 Singh
15 337 55 no pet F 34 185 Singh
16 389 87 no pet M 27 172 Campbell
17 376 78 no pet F 33 167 Campbell
18 338 57 no pet F 47 155 Li
19 337 62 no pet M 50 172 Li
20 359 69 no pet M 18 173 Li
> model<-lm(resp.flow~treatment+sex+age+height+body.weight)
> anova(model)
Analysis of Variance Table
Response: resp.flow
Df Sum Sq Mean Sq F value Pr(>F)
Treatment 1 1.8 1.8 0.0569 0.8149
sex 1 6626.9 6626.9 209.6419 8.117e-10 ***
age 1 10264.8 10264.8 324.7247 4.396e-11 ***
height 1 8346.9 8346.9 264.0536 1.758e-10 ***
body.weight 1 5786.0 5786.0 183.0373 1.978e-09 ***
Residuals 14 442.6 31.6
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
a) Identify three aspects of the design and/or analysis of this study that reduce your confidence in the conclusion that pets dont influence respiratory flow.
b) List four assumptions of the model used in this analysis.
c) If the assumptions aren’t met, what are three approaches you can take to analyse the data?
d) Which variable in their analysis explained the most variance in respiratory flow?
e) In the ANOVA table below, which letters (A-E) represent numbers that would NOT change if the sample size of people increased? (1 mark)
Response: resp.flow
Df Sum Sq Mean Sq F value Pr(>F)
Treatment A B C D E
sex 1 6626.9 6626.9 209.6419 8.117e-10 ***
age 1 10264.8 10264.8 324.7247 4.396e-11 ***
height 1 8346.9 8346.9 264.0536 1.758e-10 ***
body.weight 1 5786.0 5786.0 183.0373 1.978e-09 ***
Residuals 14 442.6 31.6
In: Math
Spring_Valley AU_Park 1375 910 1399 935 1450 1160 1270 800 970 910 1350 1020 925 1020 875 860 1000 850 1120 873 1130 1300 1200 1100 830 795 1300 1220 1220 985 925 1060 885 1040 1560 925 1380 1450 1250 1350 900 1280 900 1160 1150 975 1440 950 930 795
A) Calculate the summary statistics for both Spring Valley and AU Park. Enter the values that you calculate into the table below. Report the values to 2 decimal places.
Summary Statistics:
|
Column |
n |
Mean |
Std.dev. |
Median |
|
Spring Valley |
||||
|
AU Park |
B) What can you infer about the shape of the distribution of Spring Valley house prices by looking at the summary statistics you calculated in part c.
C) is there a difference in house prices in Spring Valley versus AU Park at α = 0.05.
(i) State the hypothesis that you want to test.
(ii) Record the value of the test statistic and its p-value.
(iii) What do you conclude for the test at α = 0.05?
D) Compute a 95% confidence interval for the difference in the mean house prices between Spring Valley and AU Park. Interpret your interval.
In: Math
a recent study in compare the time spent together by a
single and dual earner couples was 61 minutes per day, with a
standard deviation of 15.5 minutes. for the Dual earner couples the
mean number of minutes spent watching television was 48.4 minutes
and a standard deviation of 18.1 minutes. at the 0.01 significant
level we can conclude that the single earner couples on the average
spend more time watching television together? there are 15 single
earned and 12 dual earner couples studied. for calculation assume
the single earner as the first sample
In: Math
The random variable X ~ B(5, 0.8). Find the following using a
Ti-83/84
a. Mean
b. Standard Deviation
c. P(3)
d. P(x > 3)
In: Math
The Survey of Study Habits and Attitudes (SSHA) is a phychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher who suspects that older students have better attitudes toward school gives that the SSHA to 25 students who are at least 30 years of age. Their mean is ¯x = 127.8.
(a) Assuming that σ = 30 for the population of older students, carry out a test of
H0 : µ = 115
Ha : µ > 115
Report the P-value of your test, and state your conclusion clearly.
(b) Your test in part (a) required two important assumptions in addition to the assumption that the value of σ is known. What are they? Which of these assumptions is most important to the validity of your conclusion in part (a)?
In: Math
In a study of the effectiveness of a new pain killer, 46 out of 821 patients tested reported experiencing side effects. Use a a = 0.01 significance level to determine if the proportion who side effects from this drug is lower than the 7.8% rate of side effects for the older version of this medication.
In: Math
I want to test the hypothesis that a die is fair by rolling it over and over, independently, until the third time I see any single number. I will conclude that the die is loaded (not fair) if it takes four or fewer rolls for any single number to come up three times.
What is the significance level?
In: Math
Respond to the following in a minimum of 175 words, please type response:
Describe statistical inferences about two populations: How can comparisons be made for independent and dependent samples.
Respond to the following in a minimum of 175 words, please type response:
Describe analysis of variance & design of experiments: How does ANOVA extend the hypothesis testing analysis for two sample & more.
In: Math
Respond to the following in a minimum of 175 words, please type response:
An important part of using statistics is being able to explain your results to decision makers. Imagine that you have conducted a two-sample test and determined that the difference was not statistically significant. While one mean was 4.3 and the other was 3.9, the p level for the t test was p=.07. Your management team says, “Well, the difference may not be statistically significant, but the difference is there! Discuss how you would respond and how you would explain the purpose of the t test and significance in this case.
In: Math
A hotel wanted to develop a new system for delivering room service breakfasts. In the current system, an order form is left on the bed in each room. If the customer wishes to receive a room service breakfast, he or she places the order form on the doorknob before 11p.m. The current system requires customers to select a 15-minute interval for desired delivery time (6:30~6:45a.m., 6:45~7:00a.m., etc.). The new system is designed to allow the customer to request a specific delivery time. The hotel wants to measure the difference (in min.) between the actual delivery time and the requested delivery time of room service orders for breakfast (negative time means that the order was delivered before the requested time, whereas the positive time means that the order was delivered after the requested time). The factor included were the menu choice (American and Continental) and the desired time period in which the order was to be delivered (Early Time Period [6:30~8:00a.m.] or Late Time Period[8:00~9:30a.m.]). Ten orders for each combination of menu choice and desired time period were studied on a particular day, and the data were stored (BreakFast.xlsx)
(a) At the 0.05 level of significance, is there an interaction between type of breakfast and desired time?
(b) Draw the plot of means.
(c) At the 0.05 level of significance, is there an effect due to
type of breakfast? (d) At the 0.05 level of significance, is there
an effect due to desired time?
| Type | Early | Late |
| Continental | 1.4 | -2.0 |
| Continental | 1.8 | 3.1 |
| Continental | 3.3 | -0.5 |
| Continental | 4.4 | 0.9 |
| Continental | 4.4 | 1.2 |
| Continental | 5.3 | -0.1 |
| Continental | 2.2 | -1.3 |
| Continental | 1.2 | 0.2 |
| Continental | 5.4 | -0.5 |
| Continental | 1.4 | 3.8 |
| American | 4.4 | 6.0 |
| American | 1.1 | 2.3 |
| American | 4.8 | 4.2 |
| American | 8.4 | 3.8 |
| American | 6.7 | 5.8 |
| American | 5.6 | 1.8 |
| American | 9.5 | 5.1 |
| American | 5.1 | 4.2 |
| American | 7.6 | 4.9 |
| American | 9.3 | 4.2 |
In: Math