answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
love history [14]
2 years ago
10

A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next

471 orders are processed on the same day, would this prove that they are exceeding their goal, using α = .025? (a) H0: π ≤ .90 versus H1: π > .90. Choose the right option. Reject H0 if zcalc > 1.96 Reject H0 if zcalc < 1.96 a b (b) Calculate the test statistic. (Round your answer to 3 decimal places.) Test statistic 1.013 (c-1) The null hypothesis should be rejected. No Yes (c-2) The true proportion is greater than .90. No evidence to support Yes (c-3) The company is exceeding its goal. No evidence to support Yes
Mathematics
1 answer:
Kamila [148]2 years ago
8 0

Answer:

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = <u><em>proportion of orders processing on the same day they are received.</em></u>

SO, Null Hypothesis, H_0 : p \leq 0.90     {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, H_0 : p > 0.90      {means that they are exceeding their goal of 90%}

The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;

                            T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of orders that are processed on the same day = \frac{434}{471} = 0.92

           n = sample of orders = 471

So, <u><em>the test statistics</em></u>  =  \frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }

                                     =  1.599

The value of z test statistics is 1.599.

<u>Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.</u>

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

You might be interested in
a game consists of randomly choosing a bag labelled 1 2 or 3 out of a choice of 100 and then again randomly picking a ball(red o
s344n2d4d5 [400]

Answer:

.44

Step-by-step explanation:

Follow the path for each red

Add the  probabilities for each one

Bag 1 = .35*.6

Bag 2 = .45*.2

Bag 3 = .2*.7

P(red) = .35*.6 + .45* .2 + .2 *.7

          = .44

7 0
2 years ago
Read 2 more answers
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
kipiarov [429]

Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

P(k\leq11)=\sum_0^{11} P(k

P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\

P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\

P(k\leq11)=\sum_0^{11} P(k

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))

P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\

P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777

Between 11 and 23 exclusive:

P(11< k

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

P(15

P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\

3 0
2 years ago
Mike and Kim invest $18,000 in equipment to print yearbooks for schools. Each yearbook costs $5 to print and sells for $30. How
vampirchik [111]
This is a simple equation since they get 30$ and it costs  5$ that means they make 25$ each yearbook. Then you divide 18,000 by 25 and get 720

They have to sell 720 yearbooks :) hope this helped
4 0
2 years ago
Read 2 more answers
A museum employee surveys a random sample of 350 visitors to the museum. If those visitors, 266 stopped at the gift shop. Based
Free_Kalibri [48]

Answer:

about 2432 people(well exactly this many are supposed to go)

Step-by-step explanation:

266 visitors out of 350 means 19/25 people go to the gift shop. now we multiply this by 3200 to get the number of people who would go to get 2432

─────────▀▀▀▀▀▀──────────▀▀▀▀▀▀▀

──────▀▀▀▀▀▀▀▀▀▀▀▀▀───▀▀▀▀▀▀▀▀▀▀▀▀▀

────▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀──────────▀▀▀

───▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀──────────────▀▀

──▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀──────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀▀───▀▀▀▀▀▀▀───────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀─────▀▀▀▀▀▀▀──────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀▀───▀▀▀▀▀▀▀▀──────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀

─▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀

──▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀

───▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀▀

─────▀▀▀▀▀▀▀▀▀▀▀▀▀───────────────▀▀▀

──────▀▀▀▀▀▀▀▀▀▀▀───▀▀▀────────▀▀▀

────────▀▀▀▀▀▀▀▀▀──▀▀▀▀▀────▀▀▀▀

───────────▀▀▀▀▀▀───▀▀▀───▀▀▀▀

─────────────▀▀▀▀▀─────▀▀▀▀

────────────────▀▀▀──▀▀▀▀

──────────────────▀▀▀▀

───────────────────▀▀

8 0
2 years ago
Read 2 more answers
The manufacturer of a refrigerator system for beer kegs produces refrigerators that are supposed to maintain a mean temperature
Bingel [31]

Answer:

There is sufficient evidence to support the claim that the mean temperature is different from 45 degrees

Step-by-step explanation:

Given : Claim : The mean temperature of the refrigerators is different from 45 degrees

To Find :Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null​ hypothesis, state the conclusion in nontechnical terms.

Solution:

Claim : The mean temperature of the refrigerators is different from 45 degrees

So, Null hypothesis : H_0:\mu = 45

Alternate hypothesis :   H_a:\mu \neq 45

We are given that the conclusion is to reject the null​ hypothesis.

So, we can say that the claim is true that mean temperature is different from 45 degrees.

There is sufficient evidence to support the claim that the mean temperature is different from 45 degrees

4 0
2 years ago
Other questions:
  • Cindy owns a rectangular lot that is 18 yards long and 45 feet wide. She is going to cover the lot with square pieces of sod. If
    7·2 answers
  • A nationwide survey of 100 boys and 50 girls in the first grade found that the daily average number of boys who are absent from
    6·2 answers
  • Manuel has $600 in a savings account at the beginning of the summer. He wants to have at least $300 in the account at the end of
    8·2 answers
  • What is the completely factored form of 2x2 – 32?
    5·2 answers
  • What addition doubles fact can help you find 4+3? Explain how you know?
    7·1 answer
  • Of the 1,500 muffins a bakery sold in one day, 1/3 were chocolate. how many chocolate muffins did the bakery sell?
    6·2 answers
  • Daniel Potter bought a new car for $20,000.00. Two years later, he wanted to sell it. He was offered $14,650.00 for it. If he so
    11·2 answers
  • Let h(x)=505.5+8e−0.9x . What is h(5) ? rounded to the nearest tenth. any help would be great
    13·2 answers
  • Which describes how to graph h (x) = negative RootIndex StartRoot x + 8 EndRoot by transforming the parent function?
    6·1 answer
  • Please help I’ll give brainliest
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!