Sample mean: x̄ = 48.74
Sample standard deviation: s = 32.5857
Size of your sample: n = 50
For a 99% confidence interval: Point estimate =
In: Statistics and Probability
Even within a particular chain of hotels, lodging during the summer months can vary substantially depending on the type of room and the amenities offered. Suppose that we randomly select 50 billing statements from each of the computer databases of the Hotel A, the Hotel B, and the Hotel C chains, and record the nightly room rates. The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table.
Hotel A Hotel B Hotel C
Sample average ($) 145 160. 125
Sample standard deviation 17.6 22.6. 12.5
(a) Find a 95% confidence interval for the difference in the average room rates for the Hotel A and the Hotel C chains. (Round your answers to two decimal places.)
In: Statistics and Probability
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 410.0 gram setting. Based on a 46 bag sample where the mean is 412.0 grams, is there sufficient evidence at the 0.1 level that the bags are overfilled? Assume the standard deviation is known to be 20.0.
Step 1 of 5:
Enter the hypotheses:
Step 2 of 5:
Enter the value of the z test statistic. Round your answer to two decimal places.
Step 3 of 5:
Specify if the test is one-tailed or two-tailed.
Step 4 of 5:
Enter the decision rule.
Step 5 of 5:
Enter the conclusion.
In: Statistics and Probability
A health organization collects data on an individuals daily intake of selected nutrients and on their income level. The protein intake (in grams) over a 24hr period was determined for independent random samples random samples of 40 adults with income below the poverty level and 50 adults with income above the poverty level. The results:
Income Level | Sample Mean Protein intake (grams) | Sample Stardard Deviation | Number of Adults | |
Above Poverty (A) | 73.2 | 15.5 | 50 | |
Above Poverty (B) | 65.7 | 15.0 | 40 |
a) Complete the following to construct a 95% confidence interval for the amount by which true mean protein intake for adults having income above the poverty level exceeds that of adults having income below the poverty level
Parameter of interest: _____________
Degrees of freedom:
Standard error (to 5 decimal places):
Critical value t*:
Confidence interval (to 3 decimal places):
Conclusion:
b) Assuming that all of the results remained unchanged, for what level of confidence C would a margin of error of 5 grams be obtained when calculating a symmetric 100(1-α)% confidence interval? Express your final answer to 2 decimal places.
In: Statistics and Probability
12.26 Constructing an ANOVA table Refer to Exercise 12.5 (page 655). Using the table of group means and standard deviations, construct an ANOVA table similar to that on page 662. Based on the F statistic and degrees of freedom, compute the P-value. What do you conclude?
This is the question that is referred to...
Joystick types. Example 12.1 (page 645) describes a study designed to compare different joystick types on the time it takes to complete a navigation mission. In Exercise 12.3 (page 653), you described the ANOVA model for this study. The three joystick types are designated 1, 2, and 3. The following table summarizes the time (seconds) data.
Joystick | x̅ | s | n |
---|---|---|---|
1 | 279 | 78 | 20 |
2 | 245 | 68 | 20 |
3 | 258 | 80 | 20 |
In: Statistics and Probability
Andy Dwyer takes an exam for his women's studies class at Pawnee Community College. The exam has 20 multiple choice questions, each with 5 answer choices. If he simply guesses on each question, which of the following represents the probability that he gets 3 or fewer questions correct?
Select one:
None of these, this does not fit a binomial model.
b(20,0.8,0) + b(20,0.8,1) + b(20,0.8,2) + b(20,0.8,3)
b(20,0.2,0) + b(20,0.2,1) + b(20,0.2,2)
b(20,0.2,0) + b(20,0.2,1) + b(20,0.2,2) + b(20,0.2,3)
b(20,0.8,0) + b(20,0.8,1) + b(20,0.8,2)
In: Statistics and Probability
Many people find that they spend a lot of time on their smartphones each day without realizing how long they are scrolling through apps or playing games. A researcher wanted to test whether a mindfulness intervention would help people be more aware of how much time they spend on their phones, and if this might lead to decreased time spent on their phones each day. To perform this study, participants used an app that tracks phone usage in minutes the week before the mindfulness intervention, and the week after the intervention. Hypothetical data (in minutes) are provided in the table
before | after |
160 | 153 |
125 | 124 |
142 | 136 |
187 | 167 |
1.) What are the populations in this study?
2.) What is the mean of the Distribution of Means of Difference Scores?
3.) Calculate the t-statistic at a p level of 0.05 and identify the critical t-values
4.) What is the 95 percent Confidence Interval?
In: Statistics and Probability
2. Anna reads that the average price of regular gas in the country is $2.27 per gallon. To see if the average price of gas is the same in her state (North Carolina) as it is in the country overall, she randomly selects 32 gas stations throughout her state and records the price per gallon of regular gas at each station and finds a mean price of $2.21 and a standard deviation of $0.23.
(a) Does the data provide convincing evidence to refute the claim that the average price of gas in Anna’s state is the same as it is in the country as a whole? Perform the appropriate statistical test at the 1% level of significance.
(b) Suppose Anna had used a sample twice the size but
obtained the same summary statistics and used a 5% level of
significance instead. Would her conclusion change, and if so, how?
Explain.
In: Statistics and Probability
7. SPSS Interpretation Questions. The following SPSS output tables show the results from a new study of drug use in Boston. The study is based on a random sample of individuals (total N=250).
A. The research team’s main research question was determining the effect of diversion program for drug offenders on their frequency of drug use. Participants in the program were not arrested for their crimes, but instead diverted for drug treatment. The researchers have information on a sample of 100 participants in the program and 150 non-participants in the program. They show the descriptive statistics for heroin usage for each group as well as the results of their difference-in-means hypothesis test in the tables below.
(Please note that the information about the equality of variances test has been removed from the table above. These are usually displayed in the first few boxes. They are not needed here.)
Group Statistics
N Mean Std. deviation Std. Error Mean
no 150 12.80 13.86609 1.13216
yes 100 8.1500 12.02302 1.20230
Independent Samples Test
t= 2.737, df=248, sig (2 tailed)=.007, mean difference 4.65000, std error difference= 1.69912, 95% CI of the Difference lower: 1.30345, upper: 7.99655
i Based on these tables above, please write a brief summary of the results. Support all conclusions about the substance of the difference and statistical significance with the relevant statistics. (Note: You only need about 3 sentences here.)
ii. Using the table above, what is the exact probability of falsely rejecting the null hypothesis? In other words, what is the probability of observing the difference between means reported in the table and the null hypothesis being true? The table provides this information as one of the reported statistics.
In: Statistics and Probability
Only 4 in 10 registered voters believe that President Donald Trump is fit for office. While 86 percent of self-identified Republicans said Trump is fit to serve, an overwhelming 93 percent of Democrats said the OPPOSITE. The survey of 1577 voters was conducted. It has a margin of error of plus or minus 3%.
b. Construct the 95% CI estimate of the population proportion of registered voters who believe the President is fit for office.
c. Explain why NEITHER the 86 percent of Republicans NOR 7 percent of Democrats who believe Donald Trump is fit for office is a good estimate of the population proportion of ALL registered voters who believe the President is fit for office.
e. Compute the probability that i) 5 of 10 randomly selected Republicans believe that the President is fit for office.
ii) the probability that 5 randomly selected Democrats believe that the President is fit for office. Use the probability estimate provided.
In: Statistics and Probability
In: Statistics and Probability
The beer store sells potato chips and has does so for a long time. Over the last four years, it has observed that its weekly demand for potato chips varies quite a bit from week to week, but does not follow any seasonal or predictable pattern. The volume of potato chips is much higher than beer, and the store estimates that weekly demand can be modeled as being normally distributed with a mean of 35 bags per week, and a standard deviation of 7.
Assume that the store has a given amount of inventory at the start of a week, and has no opportunity to replenish this inventory during the week. In particular for each of the question we’ll assume that the store has 34 bags of inventory at the start of the week.
Part A:
1. Based on the assumed normal distribution for demand, what is the probability that weekly demand is less than or equal to 30 bags? (rounded to three decimal places)
2. What is the probability that weekly demand exceeds 45 bags? (rounded to three decimal places)
Part B
1. What is the probability that it will sell all 34 bags? (rounded to three decimal places)
2. What is the probability that it has at least 2 bags of chips leftover at the end of the week? (rounded to three decimal places)
3. What is the probability that unmet demand equals or exceeds 3 bags? (rounded to three decimal places)
4. What is the expected number of bags that the store will sell? (rounded to three decimal places)
5. What is the expected number of bags in inventory at the end of the week? (rounded to three decimal places)
6. What is the expected unmet demand? (rounded to three decimal places)
In: Statistics and Probability
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken seven blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.85 mg/dl.
(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error?
lower limit
upper limit
margin of error
(b) What conditions are necessary for your calculations?
σ is known
uniform distribution of uric acid
σ is unknown
normal distribution of uric acid
n is large
(c) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.04 for the mean concentration of uric acid in this patient's blood.
In: Statistics and Probability
A lock consists of 2 dials, where each dial has 6 letters. What is the probability of guessing the right combination in one try?
In: Statistics and Probability
What sample size is needed to estimate the population proportion within 1 percent using a 99 percent confidence level?
In: Statistics and Probability