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Dear prettyrini1,

Thanks for using JustAnswer.com! Here's the answer to the question that you referenced! These questions were answered on the date you mentioned by abozer. Here are his answers.

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Qustion #1:

Whirlpool, General Electric, and Bosch-Siemens

Total Sales = 8,950 + 6,350 + 3,200 = $18,500

Share = 18,500/27,060 = 0.6837 = 68.37%

Comparing the annual sales the above three manufacturers are the top.
Market share is the total sales of these three manufacturers with respect to the total sales of all manufacturers.

Qustion #2:

Part A
Range of the data is 10.1.
Standard deviation of data is 2.835.
We would expect 95.45% of the data to fall between 2 standard deviations from the mean (27.219 - 38.559).
Also we would expect 50% of the data to fall between the 1st quartile and the 3rd
quartile (30.600 - 34.500).
1st quartile tells us that 25% percent of the data falls below 30.600. Median tells us that 50% of the data falls below 32.300. 3rd quartile tells us that 75% of the data falls below 34.500.

Part B
Data is skewed positively. More of the data points lie below the mean.

Part C
On the average our ford cars can go 32.889 miles per gallon. While a minimum of 29.1 or a maximum of 39.2 miles per gallon can be observed on different roads.

Question #3:

Part a

Probability(Midwest or South) = (968 + 198) / 1976 = 0.590

Part b

Probability(Physician's visit given that it is from the East) = 233 / 408 = 0.571

Question #4:

P(at least 7) = P(7) + P(8) + P(9) + P(10)
= 0.11719 + 0.04395 + 0.00977 + 0.00098
= 0.17189

Question #5:

part a

P(13 < x < 20) = P(x < 20) - P(x < 13)
= 0.9719 - 0.1016
= 0.8703 = 87.03%

part b

P(x < 10) = 0.0042 = 0.42%

1000 x 0.0042 = 4 cars

Question #6:

P(z < 1.28) = 0.90
z = 1.28 = (x - 42.8)/6.2
x = 50.75 minutes

Question #7:

part a

99% confidence interval: (2.7%, 12.3%)
The actual percentage of the listeners of KGSM are between 2.7% and 12.3% of the market.

part b

We never know the actual share but we can say according to the result of the survey that it is between 2.7% and 12.3% having an average of 8% with a probability of 0.99. The salesperson is not correct!

Question #8:
part a

95% confidence interval: (1949.484, 2290.516)
The mean lifetime of GE fluroscent bulbs is between 1949.484 hours and 2290.516 hours with a probability of 0.95.

part b

We never know the actual lifetime of the bulbs but according to the sample we can say that the mean lifetime is 2120 hours with a 95% confidence interval of (1949.484, 2290.516). GE executive is not correct! The real mean could be 2200 hours but he can not give an exact value without measuring all of the lifetimes of all the GE bulbs, which is actually impossible.

Question #9:

Ho: p = 0.75 (null hypothesis)
Ha: p < 0.75 (alternative hypothesis)

p-value = 0.0362

We reject the null hypothesis since the p-value is less than the significance level (0.05).

We have sufficient evidence to conclude that less than 75% of the subscribers renew their subscription.

Question #10:

Ho: µ = 8 (null hypothesis)
Ha: µ < 8 (alternative hypothesis)

p-value = 0.0025

We reject the null hypothesis since the p-value is less than the significance level (0.01).

We have sufficient evidence to conclude that the new camera can not take an average of at least 8 photos per second.

Question #11:

part a

y = 16.0089 + 8.2324x

part b

b0: There is a fixed cost of $16.0089 even when there is no production.
b1: Labor cost increases by $8.2324 for a unit increase in batch size.

part c

r^2 = 0.994
99.4% of the variation in labor cost can be explained by the variation in batch size.

part d

r = 0.997
There is a strong positive correlation between the labor cost and the batch size.

part e

p-value = 1.01x10^-4
p-value is less than 0.05 so this data provide significant evidence that the direct labor costs are associated with the size of a batch.
Probability of obtaining a result as extreme as this by only chance is 1.01x10^-4.

part f

y = 16.0089 + 8.2324(100) = 839.254

part g

95% confidence interval: (816.058, 862.450)
For a batch size of 100, labor cost is expected to be between $816.06 and $862.45, 95% of the time.

Question #12:

part a
Demand (in hundreds of thousands of bottles) = 8.1495 - 2.1505(price (in dollars)) + 1.4409(avg. industry price (in dollars)) + 0.4008(advertising expenditure (in hundreds of thousands of dollars))

part b
p-value of the F-test is 1.74x10^-6
The model is statistically significant since the p-value is very small.

part c

This data provides significant evidence that demand (in hundreds of thousands of bottles) is associated with price (in dollars) since the p-value (1.72x10^-6) is less than the significance level (0.05).

This data provides significant evidence that demand (in hundreds of thousands of bottles) is associated with the avg. industry price (in dollars) since the p-value (5.58x10^-6) is less than the significance level (0.05).

This data provides significant evidence that demand (in hundreds of thousands of bottles) is associated with the advertising expenditure (in hundreds of thousands of dollars) since the p-value (0.0008) is less than the significance level (0.05).

part d
using the table above, predict the demand when the price is $3.70, the avg. industry price is $3.90, and the advertising expenditure is $650,000.
Demand (in hundreds of thousands of bottles) = 8.1495 - 2.1505(price (in dollars)) + 1.4409(avg. industry price (in dollars)) + 0.4008(advertising expenditure (in hundreds of thousands of dollars))
Demand (in hundreds of thousands of bottles) = 8.1495 - 2.1505(3.7) + 1.4409(3.9) + 0.4008(6.5) = 8.41737
= 841,737 bottles

Edited by Greg Michaelson on 12/21/2008 at 12:42 PM

Expert:	Greg
Pos. Feedback:	99.6 %
Accepts:
Answered:	12/20/2008

Graduate Student

I teach math and statistics at a college level as a graduate teaching assistant