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Gre4nikov [31]
3 years ago
5

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

et. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 11 sales receipts for mail-order sales results in a mean sale amount of $79.00 with a standard deviation of $18.25. A random sample of 18 sales receipts for internet sales results in a mean sale amount of $96.70 with a standard deviation of $20.25. Using this data, find the 98% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed.
Mathematics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

−35.713332 ; 0.313332

Step-by-step explanation:

Given that:

Sample size, n1 = 11

Sample mean, x1 = 79

Standard deviation, s1 = 18.25

Sample size, n2 = 18

Sample mean, x2 = 96.70

Standard deviation, s2 = 20.25

df = n1 + n2 - 2 ; 11 + 18 - 2 = 27

Tcritical = T0.01, 27 = 2.473

S = sqrt[(s1²/n1) + (s2²/n2)]

S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]

S = 7.284

(μ1 - μ2) = (x1 - x2) ± Tcritical * S

(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284

(μ1 - μ2) = - 17.7 ± 18.013332

-17.7 - 18.013332 ; - 17.7 + 18.013332

−35.713332 ; 0.313332

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−3(y+5)=15 someone please help me with this question my assignment is already late
AnnyKZ [126]

Answer: -10

Step-by-step explanation:

Simplifying

-3(y + 5) = 15

Reorder the terms:

-3(5 + y) = 15

(5 * -3 + y * -3) = 15

(-15 + -3y) = 15

Solving

-15 + -3y = 15

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '15' to each side of the equation.

-15 + 15 + -3y = 15 + 15

Combine like terms: -15 + 15 = 0

0 + -3y = 15 + 15

-3y = 15 + 15

Combine like terms: 15 + 15 = 30

-3y = 30

Divide each side by '-3'.

y = -10

Simplifying

3 0
3 years ago
What is the factored form of x2 - 6x – 16?
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x^2 - 6x - 16
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At the school carnival, the sixth graders are making directional arrows, as shown below.
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Making high-stakes insurance decisions. The Journal of Economic Psychology (Sept. 2008) published the results of a high-stakes e
Andrej [43]

Answer:

a. P(24)=0.00007

b. P(23)=0.00018

c. There is significant difference between the probability of the rainy days and the probabilities of fire and theft.

The probability of theft would be overestimated by 76% and the probability of fire would be subestimated by 27%.

Step-by-step explanation:

The probabilities of two events ("fire"and "theft") are compared to the probabilities of a certain number of days of rain during July.

The probabilities of "fire"and "theft" are around P=0.0001, and we need to calculate if the probability of exactly 23 and exactly 24 days of rain July have approximately the same probability.

Rain frequencies for the months of July and August were shown to follow a Poisson distribution with a mean of 10 days per month.

The parameter then is:

\lambda=10

The probability of k days of rain is:

P(k)=\frac{10^ke^{-10}}{k!}

For 24 days, the probability is:

P(24)=\frac{10^{24}e^{-10}}{24!}=\frac{1*10^{24}*4.54*10^{-5}}{6.20*10^{23}}  = 0.00007

The probability of 23 days of rain is 27% less than P=0.0001.

For 23 days of rain, the probability is:

P(23)=\frac{10^{23}e^{-10}}{23!}=\frac{1*10^{23}*4.54*10^{-5}}{2.59*10^{23}}  = 0.00018

The probability of 23 days of rain is 76% more than P=0.0001.

There is significant difference between the probability of the rainy days and the probabilities of fire and theft.

The probability of theft would be overestimated by 76% and the probability of fire would be subestimated by 27%.

8 0
3 years ago
he port of South Louisiana, located along miles of the Mississippi River between New Orleans and Baton Rouge, is the largest bul
iVinArrow [24]

Answer:

a) P(x < 5) = 0.7291

b) P(x ≥ 3) = 0.9664

c) P(3 < x < 4) = 0.2373

d) 5.35 million tons of cargo in a week will require the port to extend operating hours.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 4.5 million tons of cargo per week

Standard deviation = σ = 0.82 million

a) The probability that the port handles less than 5 million tons of cargo per week

= P(x < 5)

We first standardize/normalize 5.

The standardized score of any value is the value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (5 - 4.5)/0.82 = 0.61

To determine the probability that the port handles less than 5 million tons of cargo per week

P(x < 5) = P(z < 0.61)

We'll use data from the normal probability table for these probabilities

P(x < 5) = P(z < 0.61) = 0.72907 = 0.7291 to 4 d.p

b) The probability that the port handles 3 or more million tons of cargo per week?

P(x ≥ 3)

We first standardize/normalize 3.

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

To determine the probability that the port handles less than 3 or more million tons of cargo per week

P(x ≥ 3) = P(z ≥ -1.83)

We'll use data from the normal probability table for these probabilities

P(x ≥ 3) = P(z ≥ -1.83)

= 1 - P(z < -1.83)

= 1 - 0.03362

= 0.96638 = 0.9664

c) The probability that the port handles between 3 million and 4 million tons of cargo per week = P(3 < x < 4)

We first standardize/normalize 3 and 4.

For 3 million

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

For 4 million

z = (x - μ)/σ = (4 - 4.5)/0.82 = -0.61

To determine the probability that the port handles between 3 million and 4 million tons of cargo per week

P(3 < x < 4) = P(-1.83 < z < -0.61)

We'll use data from the normal probability table for these probabilities

P(3 < x < 4) = P(-1.83 < z < -0.61)

= P(z < -0.61) - P(z < -1.83)

= 0.27093 - 0.03362

= 0.23731 = 0.2373 to 4 d.p

d) Assume that 85% of the time the port can handle the weekly cargo volume without extending operating hours. What is the number of tons of cargo per week that will require the port to extend its operating hours?

Let x' represent the required number of tons of cargo thay will require the port to extend its operating hours.

Let its z-score be z'

P(x < x') = P(z < z') = 85% = 0.85

Using the normal distribution table,

z' = 1.036

z' = (x' - μ)/σ

1.036 = (x' - 4.5)/0.82

x' - 4.5 = (0.82 × 1.036) = 0.84952

x' = 0.84952 + 4.5 = 5.34952 = 5.35 to 2 d.p

Therefore, 5.35 million tons of cargo in a week will require the port to extend its operating hours.

Hope this Helps!!!

3 0
3 years ago
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