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2 years ago

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There are 1,000 golden delicious and 1,000 red delicious apples in a cooler. In a random sample of 75 of the golden delicious ap

ples, 48 had blemishes. In a random sample of 75 of the red delicious apples, 42 had blemishes. Assume all conditions for inference have been met. Which of the following is closest to the interval estimate of the difference in the numbers of apples with blemishes (golden delicious minus red delicious) at a 98 percent level of confidence?A. (–0.076,0.236)B. (–0.105,0.265)C. (–10.5,26.5)D. (–76,236)E. (−105,265)

1 answer:

Delvig [45]2 years ago

4 0

Answer:

98% confidence interval is (-0.105 , 0.265)

Step-by-step explanation:

n1 = 75 n2 = 75

X1 = 48 X2 = 42

P1 = X1/N1 = 0.64 P2 = X2/N2 = 0.56

98% confidence interval for difference between two proportions is :

P1 - P2 <u>+</u> Z $\sqrt{P1 (1-P1)/n1 + P2 (1-P2)/n1}$

= 0.08 <u>+</u> 0.185

= (-0.105 , 0.265)

98% confidence interval is (-0.105 , 0.265)

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An electrical engineer has on hand two boxes of resistors, with five resistors in each box. The resistors in the first box are l

Tju [1.3M]

Answer:

Step-by-step explanation:

given that an electrical engineer has on hand two boxes of resistors, with five resistors in each box. The resistors in the first box are labeled 10 ohms, but in fact their resistances are 9, 10, 11, 12, and 13 ohms. The resistors in the second box are labeled 20 ohms, but in fact their resistances are 18, 19, 20, 21, and 22 ohms. The EE chooses one resistor from each box and determines the resistance of each

Sample space for selecting one resistor from first box

= {9, 10, 11, 12, 13}

A = {11,12,13} (because >10)

Sample space for selecting one resistor from second box

={18,19,20,21,22}

B={18}

Sample space for sum would be

{27, 28, 29, 30, 31, 32, 33, 34, 35}

C = {(9,19) , (10,18)}

8 0

2 years ago

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un

Nataly_w [17]

Answer: 0.1289

Step-by-step explanation:

Given : The proportion of all students at a large university are absent on Mondays. : $p=0.15$

Sample size : $n=12$

Mean : $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x be a binomial variable.

Using the standard normal distribution table ,

$P(x=4)=P(x\leq4)-P(x\leq3)$ (1)

Z score fro normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Then , from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$

Hence, the probability that four students are absent = $0.1289$

3 0

2 years ago

You mix soda water, fruit punch concentrate, and ginger ale in the ratio of 1 : 2 : 5 to make fruit punch. How many pints of eac

EastWind [94]

So the ratio is 1 to 2 to 5
so basically 1 unit of soda water to 2 units of fruit punch to 5 units of ginger ale

total is 1+2+5=8 units

so 4 gallons=8 units
divide by 8
1/2 gallon=1 unit

soda water=1 unit=1/2 gallon
fruit punch=2 unit=1/2 times 2=1 gallon
ginger ale=5 unit=5 times 1/2=5/2=2 and 1/2 gallon

soda water=1/2 gallon
fruit punch concentrate=1 gallon
ginger ale=5/2 gallon or 2 and 1/2 gallon

8 0

2 years ago

29) Adam’s sister wants to watch her intake of fat per day. Here is a guideline of the recommended daily allowances. Typical val

ziro4ka [17]

Basically all the information in this problem is useless. They included all those numbers to confuse you. If you read throughly you can see that all you need to know is that Adam's sister can only have 70 grams of fat each day and that her dinner will have 48 grams of fat.

To figure this out I use this formula all the time to figure out percentages.
is/of=x/100
Your trying to figure out what is 48 percent of 70 because that's how much Adam's sister is going to eat at dinner. So your is would be 48 and your of is 70. When you substitute those numbers your formula becomes 48/70=x/100. Now you cross multiply and get 70x=4800.All that's left is to divide 70 on both sides. Which gives you 68.57.The final answer is 69 percent.

8 0

2 years ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

2 years ago