the shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. However, records indicate that the mean time is 21.2 minutes, and the standard deviation is 3.4 minutes.
Complete parts (a) through (c).
(a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required?
(b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 20 minutes?
(c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager.
In: Math
Assume a normal distribution of the form N(100, 100), answer the following questions:
What proportion of the distribution falls within 1 standard deviation of the mean (e.g., within -1 and 1 standard deviation)?
What is the probability of a single draw from that distribution has a value greater than 115?
What is the range that captures the middle 95% of the population distribution?
If I randomly sample 10 observations from this distribution and calculate a mean, what is the probability that this mean is greater than 106?
If we center the sampling distribution on 100, then what is the range that will capture 95% of the means calculated from a sample of 20 observations?
In: Math
The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data are as follows:
Pressure (psig) |
|||
Temperature (ºC) |
200 |
215 |
230 |
150 |
90.4 |
90.7 |
90.2 |
90.2 |
90.6 |
90.4 |
|
160 |
90.1 |
90.5 |
89.9 |
90.3 |
90.6 |
90.1 |
|
170 |
90.5 |
90.8 |
90.4 |
90.7 |
90.9 |
90.1 |
In: Math
Test if the population mean price for clarity “VS1” is different than that for clarity “VVS1 or VVS2”.
Please answer with R programming code
Clarity | Price |
VS2 | 1302 |
VS1 | 1510 |
VVS1 | 1510 |
VS1 | 1260 |
VS1 | 1641 |
VS1 | 1555 |
VS1 | 1427 |
VVS2 | 1427 |
VS2 | 1126 |
VS1 | 1126 |
VS1 | 1468 |
VS2 | 1202 |
VS2 | 1327 |
VS2 | 1098 |
VS1 | 1693 |
VS1 | 1551 |
VS1 | 1410 |
VS2 | 1269 |
VS1 | 1316 |
VS2 | 1222 |
VS1 | 1738 |
VS1 | 1593 |
VS1 | 1447 |
VS2 | 1255 |
VS1 | 1635 |
VVS2 | 1485 |
VS2 | 1420 |
VS1 | 1420 |
VS1 | 1911 |
VS1 | 1525 |
VS1 | 1956 |
VVS2 | 1747 |
VS1 | 1572 |
VVS2 | 2942 |
VVS2 | 2532 |
VS1 | 3501 |
VS1 | 3501 |
VVS2 | 3501 |
VS1 | 3293 |
VS1 | 3016 |
VVS2 | 3567 |
VS1 | 3205 |
VS2 | 3490 |
VS1 | 3635 |
VVS2 | 3635 |
VS1 | 3418 |
VS1 | 3921 |
VVS2 | 3701 |
VS1 | 3480 |
VVS2 | 3407 |
VS1 | 3767 |
VVS1 | 4066 |
VVS2 | 4138 |
VS1 | 3605 |
VVS2 | 3529 |
VS1 | 3667 |
VVS2 | 2892 |
VVS2 | 3651 |
VVS2 | 3773 |
VS1 | 4291 |
VVS1 | 5845 |
VVS2 | 4401 |
VVS1 | 4759 |
VVS1 | 4300 |
VS1 | 5510 |
VS1 | 5122 |
VVS2 | 5122 |
VS2 | 3861 |
VVS2 | 5881 |
VS1 | 5586 |
VS2 | 5193 |
VVS2 | 5193 |
VS2 | 5263 |
VVS2 | 5441 |
VS2 | 4948 |
VS2 | 5705 |
VS2 | 6805 |
VVS2 | 6882 |
VS1 | 6709 |
VVS2 | 6682 |
VS1 | 3501 |
VVS1 | 3432 |
VVS1 | 3851 |
IF | 3605 |
VS1 | 3900 |
VVS1 | 3415 |
IF | 4291 |
IF | 6512 |
VS1 | 5800 |
VVS1 | 6285 |
In: Math
Question 15.
What are the three required assumptions for the appropriate use of the independent groups t-test? What are the three required assumptions for the appropriate use of the dependent groups t-test?Can you use these tests when you have three groups? What test do we use instead? Can the dependent variable be nominal? What should the nature of the dependent variable be?
In: Math
A team of health research wants to investigate whether having regular lunch break improves working adults’ sleep. A group of 85 adults with a full-time job were recruited in the study, they reported average hours of sleep in the past week, committed to having a 1-hour, out-of-office, work-free lunch break each day for 3 months, and, at the end of the study, reported average hours of sleep in the past week. The mean difference in hours of sleep before and after the study was 3 with a standard deviation of 0.9. Which one of the following statements is INCORRECT? (Set alpha level at 0.05.)
The decision should be to reject the null hypothesis. |
The null hypothesis is that hours of sleep remain the same before and after the study. |
The obtained test statistic is 3.073. |
The degrees of freedom is 84. |
In: Math
We have 95 students in a class. Their abilities/eagerness are uniform randomly distributed on a scale between 1 and 4; and at the end of the class they will be judged right and they will receive a grade corresponding to their ability/eagerness (corresponding to their performance). What is the probability that the class average will be between 2.8 and 4? How would this number change (if it does) for 120 students?
In: Math
(PLEASE, SINCE THE VERY BEGINNING, ALL THE ONE BY ONE STEPS NEED TO BE MENTIONED IN YOUR CALCULATION) A manufacturer of cell phones guarantees that his cell phones will last, on average, 3 years with a standard deviation of 1 year. If 5 of those cell phones are found to have lifetimes of 1.9, 2.4, 3.0, 3.5 and 4.2 years, can the manufacturer still be convinced that his cell phones have a standard deviation of 1 year? Test at a 0.05 level of confidence. Thank you in advance for your help!
In: Math
Eyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they made the time period of January 1 to March 31. Table #11.2.4 gives the defect and the number of defects.
Table #11.2.4: Number of Defective Lenses
Defect type |
Number of defects |
Scratch |
5865 |
Right shaped - small |
4613 |
Flaked |
1992 |
Wrong axis |
1838 |
Chamfer wrong |
1596 |
Crazing, cracks |
1546 |
Wrong shape |
1485 |
Wrong PD |
1398 |
Spots and bubbles |
1371 |
Wrong height |
1130 |
Right shape - big |
1105 |
Lost in lab |
976 |
Spots/bubble - interim |
976 |
Do the data support the notion that each defect type occurs in the same proportion? Test at the 10% level.
In: Math
The types of raw materials used to construct stone tools found at an archaeological site are shown below. A random sample of 1486 stone tools were obtained from a current excavation site.
Raw Material | Regional Percent of Stone Tools | Observed Number of Tools as Current excavation Site |
Basalt | 61.3% | 905 |
Obsidian | 10.6% | 163 |
Welded Tuff | 11.4% | 165 |
Pedernal chert | 13.1% | 201 |
Other | 3.6% | 52 |
Use a 1% level of significance to test the claim that the regional distribution of raw materials fits the distribution at the current excavation site.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: The distributions are the same.
H1: The distributions are the same.
H0: The distributions are the same.
H1: The distributions are
different.
H0: The distributions are different.
H1: The distributions are different.
H0: The distributions are different.
H1: The distributions are the same.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
Yes
No
What sampling distribution will you use?
normal
Student's t
binomial
uniform
chi-square
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.100
0.050 < P-value < 0.100
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 0.01 level of significance, the evidence is sufficient to conclude that the regional distribution of raw materials does not fit the distribution at the current excavation site.
At the 0.01 level of significance, the evidence is insufficient to conclude that the regional distribution of raw materials does not fit the distribution at the current excavation site.
In: Math
Use the data in the file andy.dta consisting of data on hamburger franchises in 75 cities from Big Andy's Burger Barn.
Set up the model
ln(Si)=b1 + b2ln(Ai) + ei,
where
Si = Monthly sales revenue ($1000s) for the i-th firm
Ai = Expenditure on advertising ($1000s) for the i-th firm
(a) Interpret the estimates of slope and intercept.
(b) How well did the model fit to the data? Use any tests and measures presented in class.
(c) Perform any test for heteroscedasticity in your data.
sales | price | advert |
73.2 | 5.69 | 1.3 |
71.8 | 6.49 | 2.9 |
62.4 | 5.63 | 0.8 |
67.4 | 6.22 | 0.7 |
89.3 | 5.02 | 1.5 |
70.3 | 6.41 | 1.3 |
73.2 | 5.85 | 1.8 |
86.1 | 5.41 | 2.4 |
81 | 6.24 | 0.7 |
76.4 | 6.2 | 3 |
76.6 | 5.48 | 2.8 |
82.2 | 6.14 | 2.7 |
82.1 | 5.37 | 2.8 |
68.6 | 6.45 | 2.8 |
76.5 | 5.35 | 2.3 |
80.3 | 5.22 | 1.7 |
70.7 | 5.89 | 1.5 |
75 | 5.21 | 0.8 |
73.7 | 6 | 2.9 |
71.2 | 6.37 | 0.5 |
84.7 | 5.33 | 2.1 |
73.6 | 5.23 | 0.8 |
73.7 | 5.88 | 1.1 |
78.1 | 6.24 | 1.9 |
75.7 | 5.59 | 2.1 |
74.4 | 6.22 | 1.3 |
68.7 | 6.41 | 1.1 |
83.9 | 4.96 | 1.1 |
86.1 | 4.83 | 2.9 |
73.7 | 6.35 | 1.4 |
75.7 | 6.47 | 2.5 |
78.8 | 5.69 | 3 |
73.7 | 5.56 | 1 |
80.2 | 6.41 | 3.1 |
69.9 | 5.54 | 0.5 |
69.1 | 6.47 | 2.7 |
83.8 | 4.94 | 0.9 |
84.3 | 6.16 | 1.5 |
66 | 5.93 | 2.8 |
84.3 | 5.2 | 2.3 |
79.5 | 5.62 | 1.2 |
80.2 | 5.28 | 3.1 |
67.6 | 5.46 | 1 |
86.5 | 5.11 | 2.5 |
87.6 | 5.04 | 2.1 |
84.2 | 5.08 | 2.8 |
75.2 | 5.86 | 3.1 |
84.7 | 4.89 | 3.1 |
73.7 | 5.68 | 0.9 |
81.2 | 5.83 | 1.8 |
69 | 6.33 | 3.1 |
69.7 | 6.47 | 1.9 |
78.1 | 5.7 | 0.7 |
88 | 5.22 | 1.6 |
80.4 | 5.05 | 2.9 |
79.7 | 5.76 | 2.3 |
73.2 | 6.25 | 1.7 |
85.9 | 5.34 | 1.8 |
83.3 | 4.98 | 0.6 |
73.6 | 6.39 | 3.1 |
79.2 | 6.22 | 1.2 |
88.1 | 5.1 | 2.1 |
64.5 | 6.49 | 0.5 |
84.1 | 4.86 | 2.9 |
91.2 | 5.1 | 1.6 |
71.8 | 5.98 | 1.5 |
80.6 | 5.02 | 2 |
73.1 | 5.08 | 1.3 |
81 | 5.23 | 1.1 |
73.7 | 6.02 | 2.2 |
82.2 | 5.73 | 1.7 |
74.2 | 5.11 | 0.7 |
75.4 | 5.71 | 0.7 |
81.3 | 5.45 | 2 |
75 | 6.05 | 2.2 |
In: Math
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.7. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 5.1; H1: σ2 ≠ 5.1
Ho: σ2 = 5.1; H1: σ2 < 5.1
Ho: σ2 < 5.1; H1: σ2 = 5.1
Ho: σ2 = 5.1; H1: σ2 > 5.1
(b) Find the value of the chi-square statistic for the sample.
(Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original
distribution?
We assume a binomial population distribution.We assume a normal population distribution. We assume a uniform population distribution.We assume a exponential population distribution.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.100
0.050 < P-value < 0.100 0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.
At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.
(f) Find the requested confidence interval for the population
variance. (Round your answers to two decimal places.)
lower limit | |
upper limit |
Interpret the results in the context of the application.
We are 90% confident that σ2 lies above this interval.
We are 90% confident that σ2 lies within this interval.
We are 90% confident that σ2 lies below this interval.
We are 90% confident that σ2 lies outside this interval.
In: Math
1. A recent poll was conducted by the Gallup organization between April 2nd and April 8th, 2018. A total of 785 Facebook users living in the U.S. were selected using random digit dialing and were interviewed over the phone (either landline or cell phone). Respondents were asked the following question: “How concerned are you about invasion of privacy when using Facebook? Very concerned, somewhat concerned, not too concerned or not concerned.” 43% of those selected said “very concerned”. Suppose we want to study the proportion of all Facebook users who are “very concerned” about the invasion of privacy.
a) Define the parameter of interest in the context of the problem (include the symbol used to denote it).
b) What is the statistic in this problem (include the symbol used to denote it)?
c) Check the assumptions necessary to construct a normal-based confidence interval for the parameter of interest by verifying the appropriate conditions. Actually show that you checked these in the context of the problem.
d) Find the margin of error if we want 95% confidence in our estimate of the parameter of interest.
e) Find the 95% confidence interval for the parameter of interest.
f) In the context of the problem, give a conclusion based on the confidence interval you found in part e. Your sentence should start with the words “We are 95% confident that……”
In: Math
Hi,
Im doing my final project for quanatative analysis class. I was
tasked to create a regression analysis on Does the number of
probowl player have bearing on becoming all pro. I ran the
regression for the following data and got this output from Megastat
but I'm not sure what it tells me.
Regression Analysis | ||||||
r² | 0.006 | n | 239 | |||
r | 0.076 | k | 1 | |||
Std. Error | 0.494 | Dep. Var. | All Pro | |||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 0.3397 | 1 | 0.3397 | 1.39 | .2392 | |
Residual | 57.8193 | 237 | 0.2440 | |||
Total | 58.1590 | 238 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=237) | p-value | 95% lower | 95% upper |
Intercept | -0.1681 | 0.4981 | -0.337 | .7361 | -1.1493 | 0.8131 |
Pro Bowl | 0.5840 | 0.4950 | 1.180 | .2392 | -0.3911 | 1.5591 |
Sum of Pro Bowl Count | Sum of All Pro Count |
240 | 101 |
In: Math
The ideal (daytime) noise-level for hospitals is 45
decibels with a standard deviation of 9 db; which is to say,
this may not be true. A simple random sample of 80
hospitals at a moment during the day gives a mean noise level of 47
db. Assume that the standard deviation of noise level for all
hospitals is really 9 db. All answers to two places after the
decimal.
(a) A 99% confidence interval for the actual mean noise level in
hospitals is ( ___db, ___ db).
(b) We can be 90% confident that the actual mean noise level in
hospitals is db with a margin of error of ___ db.
(c) Unless our sample (of 81 hospitals) is among the most unusual
2% of samples, the actual mean noise level in hospitals is between
___ db and ___ db.
(d) A 99.9% confidence interval for the actual mean noise level in
hospitals is (___ db, ___ db).
(e) Assuming our sample of hospitals is among the most typical half
of such samples, the actual mean noise level in hospitals is
between ___ db and ___db.
(f) We are 95% confident that the actual mean noise level in
hospitals is db, with a margin of error of ___db.
(g) How many hospitals must we examine to have 95% confidence that
we have the margin of error to within 0.25 db?
(h) How many hospitals must we examine to have 99.9% confidence
that we have the margin of error to within 0.25 db?
In: Math