Questions
The part diameters are normally distributed. The lower tolerance limit corresponds to -2 z (minus two...

The part diameters are normally distributed. The lower tolerance limit corresponds to -2 z (minus two standard deviations below the mean). The upper tolerance limit corresponds to 3 z. What percent of parts will be out of tolerance? Describe the method you used to find the answers.

In: Statistics and Probability

An article reported the following data on oxidation-induction time (min) for various commercial oils: 88 104...

An article reported the following data on oxidation-induction time (min) for various commercial oils: 88 104 130 160 180 195 132 145 213 105 145 151 153 136 87 99 95 119 129 (a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.) s2 = min2 s = min (b) If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression. (Round your answer to three decimal places.) s2 = hr2 s = hr

In: Statistics and Probability

A certain article reported the following observations, listed in increasing order, on drill lifetime (number of...

A certain article reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. 12 14 20 23 32 37 40 43 48 51 58 61 64 66 69 70 75 77 78 79 81 83 86 90 91 93 96 100 102 105 106 106 113 119 124 135 138 141 147 157 162 169 184 206 247 264 289 323 389 512 Compute the sample median, 25% trimmed mean, 10% trimmed mean, and sample mean for the lifetime data. (Round your answers to two decimal places.) sample median 25% trimmed mean 10% trimmed mean sample mean

In: Statistics and Probability

A young investment manager tells his client that the probability of making a positive return with...

A young investment manager tells his client that the probability of making a positive return with his suggested portfolio is 94%. If it is known that returns are normally distributed with a mean of 6.2%, what is the risk, measured by standard deviation, that this investment manager assumes in his calculation? (Round "z" value to 2 decimal places and final answer to 2 decimal places. Note: if you get 7.52 % for your answer, enter it as .08. 7.52% is .0752, so rounding it to 2 decimal places gives .08)

In: Statistics and Probability

There are quarters of sales data 500,350,250,400 450,350,200,300 350,200,150,400 550,350,250,550 550,400,350,600 750,500,400,650,850 Using Minitab 18.0 Winter's...

There are quarters of sales data

500,350,250,400

450,350,200,300

350,200,150,400

550,350,250,550

550,400,350,600

750,500,400,650,850

Using Minitab 18.0 Winter's time series forecasting method forecast 3 quarters ahead

The weights are level=.2;trend=.2;seasonal=.2

In: Statistics and Probability

The May 1, 2009, issue of a certain publication reported the following home sale amounts for...

The May 1, 2009, issue of a certain publication reported the following home sale amounts for a sample of homes in Alameda, CA that were sold the previous month (1,000s of $). 589 814 580 606 354 1,290 405 535 554 681 (a) Calculate and interpret the sample mean and median. The sample mean is x = thousand dollars and the sample median is x tilde = thousand dollars. This means that the average sale price for a home in this sample was $ and that half the sales were for less than the Correct: Your answer is correct. price, while half were more than the Correct: Your answer is correct. price. (b) Suppose the 6th observation had been 985 rather than 1,290. How would the mean and median change? Changing that one value lowers the sample mean but has no effect on the sample median. Changing that one value has no effect on the sample mean but raises the sample median. Changing that one value has no effect on either the sample mean nor the sample median. Changing that one value raises the sample mean but has no effect on the sample median. Changing that one value has no effect on the sample mean but lowers the sample median. (c) Calculate a 20% trimmed mean by first trimming the two smallest and two largest observations. (Round your answer to the nearest hundred dollars.) $ (d) Calculate a 15% trimmed mean. (Round your answer to the nearest hundred dollars.) $

In: Statistics and Probability

In a survey of 3939 ​adults, 705 say they have seen a ghost. Construct a​ 99%...

In a survey of 3939 ​adults, 705 say they have seen a ghost. Construct a​ 99% confidence interval for the population proportion. Interpret the results.

A​ 99% confidence interval for the population proportion is ​( ​, ​). ​(Round to three decimal places as​ needed.) Interpret your results.

Choose the correct answer below.

A. With​ 99% confidence, it can be said that the sample proportion of adults who say they have seen a ghost is between the endpoints of the given confidence interval.

B. With​ 99% confidence, it can be said that the population proportion of adults who say they have seen a ghost is between the endpoints of the given confidence interval.

C. With​ 99% probability, the population proportion of adults who say they have not seen a ghost is between the endpoints of the given confidence interval.

D. The endpoints of the given confidence interval show that​ 99% of adults have seen a ghost.

In: Statistics and Probability

Each box of Healthy Crunch breakfast cereal contains a coupon entitling you to a free package...

Each box of Healthy Crunch breakfast cereal contains a coupon entitling you to a free package of garden seeds. At the Healthy Crunch home office, they use the weight of incoming mail to determine how many of their employees are to be assigned to collecting coupons and mailing out seed packages on a given day. (Healthy Crunch has a policy of answering all its mail on the day it is received.) Let x = weight of incoming mail and y = number of employees required to process the mail in one working day. A random sample of 8 days gave the following data.

x (lb) 14 22 15 6 12 18 23 25
y (Number of employees) 7 10 9 5 8 14 13 16

In this setting we have Σx = 135, Σy = 82, Σx2 = 2563, Σy2 = 940, and Σxy = 1530.

(f) Find Se. (Round your answer to three decimal places.)
Se =

(g) Find a 95% for the number of employees required to process mail for 14 pounds of mail. (Round your answer to two decimal places.)

lower limit     employees
upper limit     employees


(h) Test the claim that the slope β of the population least-squares line is positive at the 1% level of significance. (Round your test statistic to three decimal places.)

t =



Find or estimate the P-value of the test statistic.

P-value > 0.250

0.125 < P-value < 0.250   

0.100 < P-value < 0.125

0.075 < P-value < 0.100

0.050 < P-value < 0.075

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

0.0005 < P-value < 0.005

P-value < 0.0005


Conclusion

Reject the null hypothesis, there is sufficient evidence that β > 0.

Reject the null hypothesis, there is insufficient evidence that β > 0.   

Fail to reject the null hypothesis, there is sufficient evidence that β > 0.

Fail to reject the null hypothesis, there is insufficient evidence that β > 0.


(i) Find an 80% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit    
upper limit    


Interpretation

For each less pound of mail, the number of employees needed increases by an amount that falls within the confidence interval.

For each additional pound of mail, the number of employees needed increases by an amount that falls outside the confidence interval.    

For each additional pound of mail, the number of employees needed increases by an amount that falls within the confidence interval.

For each less pound of mail, the number of employees needed increases by an amount that falls outside the confidence interval.

In: Statistics and Probability

CASE STUDY: Chest Sizes of Scottish Militiamen (p. 306): Chest Size Frequency 33 3 34 19...

CASE STUDY: Chest Sizes of Scottish Militiamen (p. 306):

Chest Size

Frequency

33

3

34

19

35

81

36

189

37

409

38

753

39

1062

40

1082

41

935

42

646

43

313

44

168

45

50

46

18

47

3

48

1

  1. The population mean and population standard deviation of the chest circumferences are 39.85 and 2.07, respectively, identify the normal curve that should be used for the chest circumferences
  2. Use the table above to find the percentage of militiamen in the survey with chest circumference between 36 and 41 inches, inclusive. Note: as the circumferences were rounded to the nearest inch, you are actually finding the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches
  3. Use the normal curve you identified in part (a) to obtain an approximation to the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches. Compare your answer to the exact percentage found in part (b).

In: Statistics and Probability

1. Test the claim that the proportion of people who own cats is significantly different than...

1. Test the claim that the proportion of people who own cats is significantly different than 30% at the 0.2 significance level.
The null and alternative hypothesis would be:

a) H0:μ≤0.3
Ha:μ>0.3

b) H0:p≥0.3
Ha:p<0.3

c) H0:μ≥0.3
Ha:μ<0.3

d) H0:p≤0.3
Ha:p>0.3

e) H0:μ=0.3
Ha:μ≠0.3

f) H0:p=0.3
Ha:p≠0.3


The test is:

-left-tailed

-two-tailed

-right-tailed


Based on a sample of 400 people, 31% owned cats
The p-value is: ____? (to 2 decimals)


Based on this we:

  • Reject the null hypothesis
  • Fail to reject the null hypothesis

2. You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01.
      Ho:μ=89.7
      Ha:μ≠89.7
You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=12 with mean M=93.7 and a standard deviation of SD=8.6

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value = ______?
The p-value is...

  • less than (or equal to) αα
  • greater than αα

This p-value leads to a decision to...

  • reject the null
  • accept the null
  • fail to reject the null

As such, the final conclusion is that...

  • There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 89.7.
  • There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 89.7.
  • The sample data support the claim that the population mean is not equal to 89.7.
  • There is not sufficient sample evidence to support the claim that the population mean is not equal to 89.7.

In: Statistics and Probability

Suppose you have a bag of 100 coins. Ninety of them are fair coins, which is...

Suppose you have a bag of 100 coins. Ninety of them are fair coins, which is P(H) = P(T) = 1/2. The other 10 coins are biased , they have tail (T) on both sides. Let X = {X1, X2, · · · , X100} be a random variable denoting the the total number of heads in the 100 coin flips.

a. How many possible values X can take? For example, if one tosses two fair coins, the number of total heads could be 0, 1, 2.

b. How many ways the value of X could be 1?

c. How many ways the value of X could be 2?

d. How many ways the value of X could be 25?

In: Statistics and Probability

The Wind Mountain archaeological site is located in southwestern New Mexico. Wind Mountain was home to...

The Wind Mountain archaeological site is located in southwestern New Mexico. Wind Mountain was home to an ancient culture of prehistoric Native Americans called Anasazi. A random sample of excavations at Wind Mountain gave the following depths (in centimeters) from present-day surface grade to the location of significant archaeological artifacts†.

85 45 120 80 75 55 65 60
65 95 90 70 75 65 68

(b) Compute a 98% confidence interval for the mean depth μ at which archaeological artifacts from the Wind Mountain excavation site can be found. (Round your answers to one decimal place.)

lower limit       cm
upper limit       cm

In: Statistics and Probability

Describe examples of situations where you could appropriately use the following. Explain why the procedure is...

Describe examples of situations where you could appropriately use the following. Explain why the procedure is the correct one to use and identify the statistic to be used and explain why that is the correct choice. (1 point each)

  1. Confidence interval for a proportion.
  2. Confidence interval for a mean, unknown σ.
  3. Confidence interval for a mean, σ known.

Describe procedure for following and provide an example of the calculations:

  1. Margin of error calculation for 95% confidence level.
  2. Sample size determination for 95% confidence level.

In: Statistics and Probability

Why are the Maximin, Maximax, and the Minimax regret models? Not appropriate? When some conditions and...

Why are the Maximin, Maximax, and the Minimax regret models? Not appropriate? When some conditions and probabilities have a moderate level of certainty or predictability?

In: Statistics and Probability

A. As a health policy expert, you are interested in knowing whether men and women in...

A. As a health policy expert, you are interested in knowing whether men and women in the United States differ in the frequency they visit doctors when sick. So you conduct a survey of 134 people and ask them how often they visit a doctor when they are sick.  The results from this sample are as follows:

Frequency visit doctor

Sex

Always

Sometimes

Never

Total

Men

13

41

27

81

Women

24

19

10

53

Total

37

60

37

134

Perform the appropriate hypothesis test, at alpha = .01, to determine whether there is a relationship between sex and the frequency with which individuals in the United States visit doctors.  You must show your hand calculations for this problem. Describe your hypotheses, results, and conclusion in a brief paragraph.  If it is appropriate, you should also discuss the strength of the relationship based upon a suitable measure.

B . Show in a table or explain in words what the crosstabulation for your two variables from

Question above would look like if there were absolutely no association between the independent variable and the dependent variable.

In: Statistics and Probability