Questions
A lab is testing the amount of a certain active chemical compound in a particular drug...

A lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. The manufacturer claims that the average amount of the chemical is 110 mg. It is known that the standard deviation in the amount of the chemical is 7 mg.

A random sample of 21 batches of the new drug is tested and found to have a sample mean concentration of 104.5 mg of the active chemical.

a)Calculate the 95% confidence interval for the mean amount of the active chemical in the drug. Give your answers to 2 decimal places.

≤ μ ≤

b)At a significance level α = 0.05, the null hypothesis that the population mean amount of the active chemical in the drug is 110 mg is

In: Statistics and Probability

An entrepreneur examines monthly sales (in $1,000s) for 40 convenience stores in Rhode Island. (You may...

An entrepreneur examines monthly sales (in $1,000s) for 40 convenience stores in Rhode Island. (You may find it useful to reference the appropriate table: z table or t table)

Excel data

Sales Sqft
140 1810
160 2500
80 1010
180 2170
140 2310
110 1320
90 1130
110 1500
130 1950
80 1010
110 1770
140 1840
140 2330
140 2490
120 1550
120 1900
210 2320
120 1700
180 2500
170 2380
160 1880
120 1780
120 1610
90 1230
140 1920
100 1260
90 1260
190 2470
130 2420
110 1550
100 1260
140 2230
100 1500
140 1970
120 1530
120 1800
110 1520
170 2210
100 1440
110 1470

a. Select the null and the alternative hypotheses in order to test whether average sales differ from $130,000.

  • H0: μ = 130,000; HA: μ ≠ 130,000

  • H0: μ ≥ 130,000; HA: μ < 130,000

  • H0: μ ≤ 130,000; HA: μ > 130,000

b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

  • p-value < 0.01

  • 0.01 ≤ p-value < 0.02

  • 0.02 ≤ p-value < 0.05

  • 0.05 ≤ p-value < 0.10

  • p-value ≥ 0.10

c. At α = 0.05 what is your conclusion? Do average sales differ from $130,000?

  • Reject H0; average sales differ from $130,000.

  • Reject H0; average sales do not differ from $130,000.

  • Do not reject H0; average sales differ from $130,000.

  • Do not reject H0; average sales do not differ from $130,000.

In: Statistics and Probability

AlwaysRain Irrigation, Inc., would like to determine capacity requirements for the next four years. Currently two...

AlwaysRain Irrigation, Inc., would like to determine capacity requirements for the next four years. Currently two production lines are in place for making bronze and plastic sprinklers. Three types of sprinklers are available in both bronze and plastic: 90-degree nozzle sprinklers, 180-degree nozzle sprinklers, and 360-degree nozzle sprinklers. Management has forecast demand for the next four years as follows:

YEARLY DEMAND
1 (IN 000s) 2 (IN 000s) 3 (IN 000s) 4 (IN 000s)
Plastic 90 31 40 52 53
Plastic 180 14 16 17 14
Plastic 360 52 52 63 65
Bronze 90 6 6 5 15
Bronze 180 2 2 5 13
Bronze 360 10 12 11 22


Both production lines can produce all the different types of nozzles. The bronze machines needed for the bronze sprinklers require three operators and can produce up to 18,000 sprinklers. The plastic injection molding machine needed for the plastic sprinklers requires four operators and can produce up to 300,000 sprinklers. Three bronze machines and only one injection molding machine are available.

What are the capacity requirements for the next four years? (Assume that there is no learning.) (Enter the demand values in thousands. Round your answers to 2 decimal places.)


Year 1 Year 2 Year 3 Year 4
Plastic
Demand for plastic sprinklers
Percentage of capacity used % % % %
Machine requirements
Labor requirements
Bronze
Demand for bronze sprinklers
Percentage of capacity used % % % %
Machine requirements
Labor requirements

In: Statistics and Probability

How to determine quartile

How to determine quartile

In: Statistics and Probability

A manufacturing shop is designed to operate most efficiently at an output of 580 units per...

A manufacturing shop is designed to operate most efficiently at an output of 580 units per day. In the past month the plant averaged 510 units per day.

What was their capacity utilization rate last month? (Round your answer to 1 decimal place.)

Capacity utilization rate             %

In: Statistics and Probability

Question text In a random sample of 200 people, 32 people have blue eyes (characteristic B)....

Question text

In a random sample of 200 people, 32 people have blue eyes (characteristic B). Which of the following 95% confidence intervals estimate the true proportion pp of measurements in the population with characteristic B?

Select one:

a. (0.109, 0.211)

c. (0.431, 0.712)

d. (0.561, 0.912)

In: Statistics and Probability

The score of 24 randomly selected exams in a geometry class are given below: 72 85...

The score of 24 randomly selected exams in a geometry class are given below:

72 85 62 88 75 65 76 99 74 67 83 50 98 78 90 70 80 55 78 77 70 80 68 60

It has been reported that the mean score of all geometry exams is less than 78. Test the validity of the report at α = 0.02 by using the data given above.

(a) Clearly, state H0 and H1, identify the claim and type of test.

H0 :

H1 :

b) Find and name all related critical values, draw the distribution, and clearly mark and shade the critical region(s).

(c) Find the computed test statistic and the P-value.

C.T.S. :

P-Value :

(d) Use non-statistical terminology to state your final conclusion about the claim.

It has also been reported that the standard deviation of all scores in a geometry exam is 10. Test the validity of the report at α = 0.01 by using the data given above.

(e) Clearly, state H0 and H1, identify the claim and type of test.

H0 :

H1 :

(f) Find and name all related critical values, draw the distribution, clearly mark and shade the critical region(s).

(g) Find the computed test statistic and the P-value.

C.T.S. :

P-Value :

(h) Use non-statistical terminology to state your final conclusion about the claim.

In: Statistics and Probability

The average driving distance​ (yards) and driving accuracy​(percent of drives that land in the​ fairway) for...

The average driving distance​ (yards) and driving accuracy​(percent of drives that land in the​ fairway) for 8 golfers are recorded in the table to the right. Complete parts a through e below.

Player

Distance​ (yards)

Accuracy​ (%)

1

316.1316.1

44.6

2

303.3

51.7

3

309.4

47.5

4

311.6

40.6

5

295.3

55.4

6

290.4

58.9

7

295.8

56.2

8

304.3

49.5

a. Write the equation of a​ straight-line model relating driving accuracy​ (y) to driving distance​ (x). Choose the correct answer below.

A.y =β1x

B.y=β1x2+β0

C.y=β1x+ε

D.y=β0+β1x+ε

Your answer is correct.

b. Fit the​ model, part

a​,to the data using simple linear regression. Give the least squares prediction equation.ModifyingAbove y with caretyequals= 250.3plus+Left parenthesis negative 0.6587 right parenthesis(−0.6587)x

c. Interpret the estimated​ y-intercept of the line.

Choose the correct answer below.

A.Since a drive with​ 0% accuracy is outside the range of the sample​ data, the​ y-intercept has no practical interpretation.

B.For each additional yard in​ distance, the accuracy is estimated to change by the value of the​ y-intercept.

C.For each additional percentage in​ accuracy, the distance is estimated to change by the value of the​y-intercept.

D.Since a drive with distance 0 yards is outside the range of the sample​ data, the​ y-intercept has no practical interpretation.

Your answer is correct.

d. Interpret the estimated slope of the line. Choose the correct answer below.

A.Since a drive with distance 0 yards is outside the range of the sample​ data, the slope has no practical interpretation.

B.For each additional percentage in​ accuracy, the distance is estimated to change by the value of the slope.

C.For each additional yard in​ distance, the accuracy is estimated to change by the value of the slope.

D.Since a drive with​ 0% accuracy is outside the range of the sample​ data, the slope has no practical interpretation.

In: Statistics and Probability

Are America's top chief executive officers (CEOs) really worth all that money? One way to answer...

Are America's top chief executive officers (CEOs) really worth all that money? One way to answer this question is to look at row B, the annual company percentage increase in revenue, versus row A, the CEO's annual percentage salary increase in that same company. Suppose that a random sample of companies yielded the following data: B: Percent for company 2 5 29 8 21 14 13 12 A: Percent for CEO -1 5 21 13 12 18 9 8 Do these data indicate that the population mean percentage increase in corporate revenue (row B) is different from the population mean percentage increase in CEO salary? Use a 1% level of significance. Will you use a left tailed, right tailed, or two tailed test? Select one: a. two tailed test b. right tailed test c. left tailed test

In: Statistics and Probability

The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation...

The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that the average height of a sample of twenty 18-year-old men will be less than 69 inches? Round your answer to four decimal places.

In: Statistics and Probability

The (simulated) data (mg decrease per decilitre of blood) for the Omega-3 group is stored in...

The (simulated) data (mg decrease per decilitre of blood) for the Omega-3 group is stored in Table A1, and the data for the control group is stored in Table A2. You are required to calculate a 95% confidence interval for the average difference in cholesterol reduction and to test the hypothesis that there was no difference between the two diets in average reduction of cholesterol.

1. How many patients were in each diet group?

2. What was the mean (decrease) in cholesterol for the Omega-3 group of patients?

3. What was the standard deviation in that group? LABORATORY ASSIGNMENTS INTRODUCTORY STATISTICS LABORATORY 37

4. What was the mean (decrease) in cholesterol for the control group of patients?

5. What was the standard deviation in the control group?

6. What is the estimated difference of means?

7. Standard error of the difference between means 2 2 2 1 2 1 n s n s = + What is the standard error of the difference of means?

b) Calculate the margin of error of the estimated difference of means. For this large-sample 95% confidence interval we can approximate with a z value which is z0.025 = 1.96. Calculate the confidence interval as difference between means ± margin of error.

8. What is the margin of error of the estimated difference?

9. What is the lower limit for the 95% confidence interval of the difference in cholesterol reduction between Omega-3 and control diets?

10. What is the upper limit?

DATA A1:

86.6
73.9
99.4
67.9
82.3
80.3
82.8
77.9
76.8
85.9
62.5
34
120.9
57
114.9
83.7
103.5
67.4
86.1
87.6
104.2
123
91.7
95.2
95.8
90.7
72.9
72.9
62
98.1
70
77.3
34.8
58
66.7
100.6
98.6
75
52.9
54
42.1
86.7
68.9
68.8
76
87.1
99.2
119.3
88.5

90.9

DATA A2

52.5
38.6
60.7
90.8
57.9
38.6
31.9
35.2
27.7
31.5
39.4
80.4
61.1
88.1
37.5
43.5
35.6
81.6
55.2
66.7
56.3
47.4
105.2
51.7
35.8
58.9
24.5
92.8
55
52.5
110.2
86.5
30.7
50.8
58.2
75.1
73.9
12.5
23.3
96.6
55.6
102.9
103
50.2
88.7
53.7
41.5
24.7
80.3
84.4

In: Statistics and Probability

To study how social media may influence the products consumers​ buy, researchers collected the opening weekend...

To study how social media may influence the products consumers​ buy, researchers collected the opening weekend box office revenue​ (in millions of​ dollars) for 23 recent movies and the social media message rate​ (average number of messages referring to the movie per​ hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue​ (y) and message rate​ (x).

Message Rate

Revenue​ ($millions)

1363.2

146

1219.2

79

681.2

67

583.6

37

454.7

35

413.9

34

306.2

21

289.8

18

245.1

18

163.9

17

148.9

16

147.4

15

147.3

15

123.6

14

118.1

13

108.9

13

100.1

12

90.3

11

89.1

6

70.1

6

56.2

5

41.6

3

8.4

1

The least squares regression equation is ModifyingAbove y with caret=________+ ( _______ )x. ​(Round to three decimal places as​ needed.)

PrintDone

In: Statistics and Probability

1. A population of values has a normal distribution with μ=182.1 and σ=28.9. You intend to...

1. A population of values has a normal distribution with μ=182.1 and σ=28.9. You intend to draw a random sample of size n=117.

Find the probability that a single randomly selected value is less than 187.7.
P(X < 187.7) =

Find the probability that a sample of size n=117is randomly selected with a mean less than 187.7.
P(¯x < 187.7) =  

2.
CNNBC recently reported that the mean annual cost of auto insurance is 1045 dollars. Assume the standard deviation is 211 dollars. You take a simple random sample of 69 auto insurance policies.

Find the probability that a single randomly selected value is less than 977 dollars.
P(X < 977) =

Find the probability that a sample of size n=69= is randomly selected with a mean less than 977 dollars.
P(¯xx¯ < 977) =

In: Statistics and Probability

Very Bad Drugs Corp believes that their new drug, Brain Boost, increases focus and alertness by...

Very Bad Drugs Corp believes that their new drug, Brain Boost, increases focus and alertness by 15%. Answer the following:

(a)Formulate a set of two hypotheses to test their claim in words and symbols.

(b) What would a Type 1 error be?

(c) What would a Type 2 error be?

In: Statistics and Probability

Q2: Do social recommendations increase ad effectiveness? A study of online viewers who arrived at an...

Q2: Do social recommendations increase ad effectiveness? A study of online viewers who arrived at an advertising video for a particular brand by following a social media recommendation link to viewers who arrived at the same video by web browsing. Data were collected on whether the viewer could correctly recall the brand being advertised after seeing the video. The results were:

Yes No

Recommendation 407 150

Browsing 193 91

Determine whether brand recall is higher following a social media recommendation than with only web browsing at alpha=.05.

1. What is the claim from the question? What are Null and Alternative Hypothesis for this problem?

2. What kind of test do you want to use?

A. Two Samples Independent Population T Test with Equal Variance

B. Two Samples related Population T Test

C. Two Samples Independent Population T Test with Unequal Variance

D. Two Sample Proportion Z Test

E. Two Sample Variance F Test

3. Calculate Test Statistics

4. Find Critical Value(s) and appropriate degree of freedom if necessary Test Statistics:

5. Find P-value

6. What is the conclusion that you could make? Clearly write down the conclusion and business statement and illustrate what type error you could make.

In: Statistics and Probability