Questions
Suppose you toss a fair coin 10 times resulting in a sequence of heads (H) and...

Suppose you toss a fair coin 10 times resulting in a sequence of heads (H) and tails (T). Let X be the number of times that the sequence HT appears. For example, HT appears thrice in

THTHHTHHHT

Find E(X). Hint: Use Indicator random variables.

In: Statistics and Probability

A life insurance salesperson claims the average worker in the city of Rome has no more...

A life insurance salesperson claims the average worker in the city of Rome has no more than $25,000 of personal life insurance. To test this claim, you randomly sample 100 workers in Rome. You find that this sample of workers has a mean $26,650 of personal life insurance. The population standard deviation is $12,000. Determine whether the test shows enough evidence to reject the null hypothesis posed by the salesperson. Assume α = 0.05.

a) State null and alternative hypothesis

b) Determine the level of significance and identify if the test requires one-sided or two-sided test

c) Establish the decision rule using critical value, test statistic and p value method

d) Write a concluding statement

In: Statistics and Probability

Refer to the gasoline sales time series data in the given table. Week Sales (1,000s of...

Refer to the gasoline sales time series data in the given table.

Week Sales (1,000s of gallons)
1 16
2 22
3 17
4 23
5 15
6 17
7 21
8 19
9 20
10 18
11 15
12 23
(a) Compute four-week and five-week moving averages for the time series.
If required, round your answers to two decimal places.

Week

Sales
4 Period
Moving Average
5 period
Moving Average
1 16
2 22
3 17
4 23
5 15
6 17
7 21
8 19
9 20
10 18
11 15
12 23
(b) Compute the MSE for the four-week and five-week moving average forecasts.
If required, round your final answers to three decimal places.
MSE for four-week moving average =  
MSE for five-week moving average =   
(c) What appears to be the best number of weeks of past data (three, four, or five) to use in the moving average computation? Consider that the MSE for the three-week moving average is 12.667.
- Select your answer -ThreeFourFiveItem 18

In: Statistics and Probability

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according...

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)

Diet A 9 8 7 14 10 7 8 11 5 14
Diet B 8 21 20 19 17 8 10 19 11 12


(a) Find t. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

In: Statistics and Probability

3. Interval estimation of a population mean, population standard deviation unknown The Business Environment and Enterprise...

3. Interval estimation of a population mean, population standard deviation unknown

The Business Environment and Enterprise Performance Survey (BEEPS), developed by the European Bank for Reconstruction and Development and the World Bank, is a survey of more than 4,000 firms in 22 transition countries. Conducted in 2000, BEEPS gathered information on the impediments to business growth in transition countries.

As part of BEEPS, firms that import goods answered the question, “How many days does it take from the time your goods arrive in their port of entry until the time you can claim them from customs?” For the sample of 43 importing firms in Slovakia, the sample mean x̄ was 4.8 days, and the sample standard deviation s was 8.2 days.

The standard deviation of the population distribution is unknown, but you are willing to assume that the population distribution is not highly skewed and contains no outliers.

To develop a 99% confidence interval estimate of the mean number of days it takes for imported goods to clear customs in Slovakia, use the:
  • t distribution with 42 degrees of freedom
  • t distribution with 35 degrees of freedom
  • standard normal distribution
  • t distribution with 43 degrees of freedom

Use the Distributions tool to compute a 99% confidence interval estimate for the mean number of days it takes for imported goods to clear customs in Slovakia.

You are 99% confident that the mean wait time for imported goods to clear customs in Slovakia is between:
  • 1.43
  • 4.29
  • 1.78
  • 1.58
and:
  • 8.02
  • 5.31
  • 8.17
  • 7.82

The confidence interval estimate you calculated is appropriate because:

a. You are assuming the population distribution is not highly skewed nor contains outliers and the sample size is at least 30.

b. The wait time for imported goods to clear customs in Slovakia is normally distributed.

c. Using the t distribution means the sampling distribution of the mean does not need to be normal.

In: Statistics and Probability

hink about the following game: A fair coin is tossed 10 times. Each time the toss...

hink about the following game: A fair coin is tossed 10 times. Each time the toss results in heads, you receive $10; for tails, you get nothing. What is the maximum amount you would pay to play the game?

Define a success as a toss that lands on heads. Then the probability of a success is 0.5, and the expected number of successes after 10 tosses is 10(0.5) = 5. Since each success pays $10, the total amount you would expect to win is 5($10) = $50.

Suppose each person in a random sample of 1,536 adults between the ages of 22 and 55 is invited to play this game. Each person is asked the maximum amount they are willing to pay to play. (Data source: These data were adapted from Ben Mansour, Selima, Jouini, Elyes, Marin, Jean-Michel, Napp, Clotilde, & Robert, Christian. (2008). Are risk-averse agents more optimistic? A Bayesian estimation approach. Journal of Applied Econometrics, 23(6), 843–860.)

Someone is described as “risk averse” if the maximum amount he or she is willing to pay to play is less than $50, the game’s expected value. Suppose in this 1,536-person sample, 1,482 people are risk averse. Let p denote the proportion of the adult population aged 22 to 55 who are risk averse and 1 – p, the proportion of the same population who are not risk averse. Use the sample results to estimate the proportion p.

The proportion p̄p̄ of adults in the sample who are risk averse is___________ . The proportion 1 – p̄p̄ of adults in the sample who are not risk averse is ________ .

You ______ conclude that the sampling distribution of p̄p̄ can be approximated by a normal distribution, because .

The sampling distribution of p̄p̄ has an estimated standard deviation of_______ .

0123Standard Normalt Distribution

Select a Distribution

Use the Distributions tool to develop a 95% confidence interval estimate of the proportion of adults aged 22 to 55 who are risk averse.

You can be 95% confident that the interval estimate _______ to ________includes the population proportion p, the proportion of adults aged 22 to 55 who are risk averse.

In: Statistics and Probability

A local tire dealer wants to predict the number of tires sold each month. He believes...

A local tire dealer wants to predict the number of tires sold each month. He believes that the number of tires sold is a linear function of the amount of money invested in advertising. He randomly selects 6 months of data consisting of tire sales (in thousands of tires) and advertising expenditures (in thousands of dollars). Based on the data set with 6 observations, the simple linear regression model yielded the following results.

?X =24
?X2=124
?Y  = 42
?Y2 =338
?XY =196

Calculate the coefficient of determination and the coefficient of correlation between X and Y. Interpret the coefficient of Determination. also find the slope and intercept and write the estimated Regression equation. What would the predicted sales of tires be if he spends five thousand dollars in advertising? Perform the test of significance for the slope coefficient. Use 5% level of significance.

In: Statistics and Probability

If the estimated Z is 1.48, the p-value for a two-tailed test is

If the estimated Z is 1.48, the p-value for a two-tailed test is

In: Statistics and Probability

A manufacturing company for car batteries uses lead in its manufacturing process. Workers are encouraged to...

A manufacturing company for car batteries uses lead in its manufacturing process. Workers are encouraged to shower, shampoo, and change

            clothes before going home to eliminate the transfer of lead to their children, but there is still concern that children are being exposed by their parents. A study is carried out to determine whether workers carry lead dust home. 33 of the workers children were selected as subjects, and a blood tests for each child determines the level of lead in his/her blood. A control group of 33 children from the same neighborhood with no connection to the manufacturing plant is also selected, and their blood lead levels are also measured. (20 points)

Input the data into R using the following commands:

Exposed = c(38, 23, 41, 18, 37, 36, 23, 62, 31, 34, 24,

         14, 21, 17, 16, 20, 15, 10, 45, 39, 22, 35,

         49, 48, 44, 35, 43, 39, 34, 13, 73, 25, 27)

Control = c(16, 18, 18, 24, 19, 11, 10, 15, 16, 18, 18,

         13, 19, 10, 16, 16, 24, 13, 9, 14, 21, 19,

         7, 18, 19, 12, 11, 22, 25, 16, 13, 11, 13)

This creates two variables, Exposed and Control, containing the blood lead levels for the two groups, respectively. Make histograms, look at summary statistics, and look at a side-by side boxplot for the groups using the following commands:

hist(Exposed)

summary(Exposed)

hist(Control)

summary(Control)

boxplot (Exposed, Control)

what differences do you see between the sampling distributions of the two groups?

b)

            Is there evidence that the average blood lead level is different in the Exposed vs control groups? What are the relevant null and alternative hypotheses in terms of population parameters?

c)         Perform a t-test of the null hypothesis of no difference in mean with a two-sided alternative using the following command

t.test(Exposed, Control, paired=FALSE, alternative=”two.sided”)

and interpret the results.

A blood lead level of more than 45 requires medical treatment. Test the hypotheses

H0: mexp = 45

vs

H0: mexp ¹ 45

where mexp is the true mean of the exposed children, using the command

t.test(Exposed, mu=45, alternative= “two.sided”)

What is your conclusion?

e) Interpret the 95% confidence interval given in the output for part d) above.

In: Statistics and Probability

List ways that statistics is used and write an example shows why we use statistics in...

List ways that statistics is used and write an example shows why we use statistics in business.

In: Statistics and Probability

Shelia's measured glucose level one hour after a sugary drink varies according to the Normal distribution...

Shelia's measured glucose level one hour after a sugary drink varies according to the Normal distribution with µ = 118 mg/dl and s = 10 mg/dl.


What is the level L (±0.1) such that there is probability only 0.01 that the mean glucose level of 4 test results falls above L?   

In: Statistics and Probability

The------------------ is a probability value determined by the experimenter to define the critical values a. beta...

The------------------ is a probability value determined by the experimenter to define the critical values

a. beta level

b. standard error of the mean

c. observed value

d. alpha level

In: Statistics and Probability

9. In a community of 800 households (population 4320), public health authorities found 120 persons with...

9. In a community of 800 households (population 4320), public health authorities found 120 persons with condition X in 80 households. A total of 480 persons lived in the 80 affected households. Assuming that each household had only one primary case, the secondary attack rate is:

[1] [Hint: Secondary Attack Rate is calculated as New cases among contacts of primary cases during a short period of time period* 100/Population at beginning of the time period – Primary cases]. g. 8.5% h. 10.0% i. 16.7% j. 25.0% k. 30.0%

In: Statistics and Probability

The mean cost of domestic airfares in the United States rose to an all-time high of...

The mean cost of domestic airfares in the United States rose to an all-time high of $380 per ticket. Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $120. Use Table 1 in Appendix B.

a. What is the probability that a domestic airfare is $545 or more (to 4 decimals)?

b. What is the probability that a domestic airfare is $265 or less (to 4 decimals)?

c. What if the probability that a domestic airfare is between $300 and $510 (to 4 decimals)?

d. What is the cost for the 2% highest domestic airfares? (rounded to nearest dollar)

In: Statistics and Probability

Use the data and Excel to answer this question. It contains the United States Census Bureau’s...

Use the data and Excel to answer this question. It contains the United States Census Bureau’s estimates for World Population from 1950 to 2014. You will find a column of dates and a column of data on the World Population for these years. Generate the time variable t. Then run a regression with the Population data as a dependent variable and time as the dependent variable. Have Excel report the residuals.

(a) Based on the ANOVA table and t-statistics, does the regression appear significant?

(b) Calculate the Durbin-Watson Test statistic. Is there a serial correlation problem with the data? Explain.

(d) What affect might your answer in part (b) have on your conclusions in part (a)?

Year Population
1950 2,557,628,654
1951 2,594,939,877
1952 2,636,772,306
1953 2,682,053,389
1954 2,730,228,104
1955 2,782,098,943
1956 2,835,299,673
1957 2,891,349,717
1958 2,948,137,248
1959 3,000,716,593
1960 3,043,001,508
1961 3,083,966,929
1962 3,140,093,217
1963 3,209,827,882
1964 3,281,201,306
1965 3,350,425,793
1966 3,420,677,923
1967 3,490,333,715
1968 3,562,313,822
1969 3,637,159,050
1970 3,712,697,742
1971 3,790,326,948
1972 3,866,568,653
1973 3,942,096,442
1974 4,016,608,813
1975 4,089,083,233
1976 4,160,185,010
1977 4,232,084,578
1978 4,304,105,753
1979 4,379,013,942
1980 4,451,362,735
1981 4,534,410,125
1982 4,614,566,561
1983 4,695,736,743
1984 4,774,569,391
1985 4,856,462,699
1986 4,940,571,232
1987 5,027,200,492
1988 5,114,557,167
1989 5,201,440,110
1990 5,288,955,934
1991 5,371,585,922
1992 5,456,136,278
1993 5,538,268,316
1994 5,618,682,132
1995 5,699,202,985
1996 5,779,440,593
1997 5,857,972,543
1998 5,935,213,248
1999 6,012,074,922
2000 6,088,571,383
2001 6,165,219,247
2002 6,242,016,348
2003 6,318,590,956
2004 6,395,699,509
2005 6,473,044,732
2006 6,551,263,534
2007 6,629,913,759
2008 6,709,049,780
2009 6,788,214,394
2010 6,858,584,755
2011 6,935,999,491
2012 7,013,871,313
2013 7,092,128,094
2014 7,169,968,185

Thanks id advance! Will try to rate the answer ASAP. Please show your process too :)

In: Statistics and Probability