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Andrei [34K]
2 years ago
7

Two professors at a nearby university want to co-author a new textbook in either economics or statistics. They feel that if they

write an economics book they have a 50% chance of placing it with a major publisher where it should ultimately sell about 40,000 copies.
If they can’t get a major publisher to take it, then they feel they have an 80% chance of placing it with a smaller publisher, with sales of 30,000 copies.

On the other hand if they write a statistics book, they feel they have a 40% chance of placing it with a major publisher, and it should result in ultimate sales of about 50,000 copies.

If they can’t get a major publisher to take it, they feel they have a 50% chance of placing it with a smaller publisher, with ultimate sales of 35,000 copies.

What is the probability that the statistics book would wind up being placed with a smaller publisher?
Mathematics
1 answer:
WARRIOR [948]2 years ago
3 0

Answer:

0.3 or 30%

Step-by-step explanation:

The professors will only seek a smaller publisher if they cannot place the statistics book with a major publisher. If the probability of getting a major publisher is 40% and the probability of getting a smaller publisher is 50%, then the probability that the statistics book would wind up being placed with a smaller publisher is given by:

P=(1-P(major))*P(minor)\\P=(1-0.4)*0.5\\P=0.3

The probability is 0.3 or 30%.

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Jobisdone [24]

Answer: (0.760, 0.820)

Step-by-step explanation:

Let p be the population proportion of all adults feel that "education and our schools" is one of the top issues facing California.

Given : 79% (actual results are 400 out of 506 surveyed) of California adults feel that "education and our schools" is one of the top issues facing California.

i.e.Sample size : n= 506

Sample proportion : \hat{p}=0.79

Critical value for 90% confidence interval ( Using z-value table) :

z=1.645

Now, the 90% confidence interval for the population proportion will be :

\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

i.e. 0.79\pm (1.645)\sqrt{\dfrac{0.79(1-0.79)}{506}}

i.e. 0.79\pm (1.645)(0.0181)

\approx0.79\pm 0.030=(0.79-0.030,\ 0.79+0.030)\\\\=(0.760,\ 0.820)

Hence, the 90% confidence interval for the population proportion= (0.760, 0.820)

7 0
2 years ago
Solve the equation or inequality 3/2t-16=4/3t-6
Tamiku [17]

Answer:

<em>t=60</em>

Step-by-step explanation:

3/2t-16=4/3t-6

(subtract 4/3t from both sides)

3/2t-16-4/3t=-6

(add 16 to both sides)

3/2t-4/3t=-6+16

(simplify)

1/6t=10

(divide by 1/6 (or multiply by 6 on both sides)

t=60

7 0
1 year ago
From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri
Leviafan [203]

Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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