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elena-s [515]
2 years ago
8

According to a report from the United States Environmental Protection Agency, burning one gallon of gasoline typically emits abo

ut 8.9 kg of CO2. A fuel company wants to test a new type of gasoline designed to have lower CO2 emissions. Here are their hypotheses:
H0: μ = 8.9 kg
Ha: μ < 8.9 kg (where μ is the mean amount of CO2 emitted by burning one gallon of this new gasoline).

Which of the following would be a Type II error in this setting?

A. The mean amount of CO2 emitted by the new fuel is actually 89 kg but they conclude it is lower than 89 kg
B. The mean amount of CO2 emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg
C. The mean amount of CO2 emitted by the new fuel is actual 89 kg and they alto conclude it is lower than 89 kg and they conclude it is lower than 9 kg
D. The mean amount of CO2 emitted by the new fuels actually lower than 8.9
Mathematics
1 answer:
Minchanka [31]2 years ago
8 0

Answer:

B. The mean amount of CO_2 emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg

Step-by-step explanation:

A Type II error is the failure to reject a false null hypothesis.

Given the null and alternate hypothesis of a fuel company which wants to test a new type of gasoline designed to have lower CO_2 emissions.:

H_0: \mu = 8.9 kg\\H_a: \mu < 8.9 kg \\\text{ (where \mu is the mean amount of CO_2 emitted by burning one gallon of this new gasoline)}

where \mu is the mean amount of CO_2 emitted by burning one gallon of this new gasoline.

If the null hypothesis is false, then:

H_a: \mu < 8.9 kg

A rejection of the alternate hypothesis above will be a Type II error.

Therefore:

The Type II error is: (B) The mean amount of CO_2  emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg.

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Use the graph to write an explicit function to represent the data and determine how much money Amy earned in week 5.
zalisa [80]

Answer:

3rd option

f(n)=5(3)^{n-1}

f(5)=405

Step-by-step explanation:

So we are given the following points:

(1,5)

(2,15)

(3,45)

(4,135)

This is a geometric sequence because there is a common ratio, 3. That is you can keep multiply 3 to a previous y-coordinate to get the next y-coordinate.

The formula for a geometric sequence is f(n)=a_1 \cdot r^{n-1}

where a_1 is the first term and r is the common ration.

So we have

f(n)=5 \cdot (3)^{n-1}.

If you want to know the fifth term, just plug in 5:

f(5)=5 \cdot (3)^{5-1}

Simplifying:

f(5)=5 \cdot 3^4

f(5)=5 \cdot 81

f(5)=405

3 0
2 years ago
Read 2 more answers
A distribution for a set of wrist circumferences (measured in centimeters) taken from the right wrist of a random sample of newb
lozanna [386]

Answer:

A Histogram will be used to represent the size of right wrist of the random sample of newborn infants.

Step-by-step explanation:

A histogram is the graphical representation of the frequency distribution in the given sample. As the value of circumference can be a positive real number, therefore a Histogram with class boundaries can be formed such that the overall frequency of a wrist size is also visible in the graph.

Also as the distribution will be of continuous nature thus a histogram is a more suitable option as compared to a bar or stem and leaf graph.

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2 years ago
Twenty students from Sherman High School were accepted at Wallaby University. Of those students, eight were offered military sch
Shalnov [3]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let w be the subscript for scores of students with military scholarship and o be the subscript for scores of students without military scholarship.

Therefore, the population means would be μw and μo.

The random variable is xw - xo = difference in the sample mean scores of students with military scholarships and students without.

For students with military scholarship,

n = 8

Mean = (850 + 925 + 980 + 1080 + 1200 + 1220 + 1240 + 1300)/8

Mean = 1099.375

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (850 - 1099.375)^2 + (925 - 1099.375)^2 + (980 - 1099.375)^2 + (1080 - 1099.375)^2 + (1200 - 1099.375)^2 + (1220 - 1099.375)^2 + (1240 - 1099.375)^2 + (1300 -1099.375)^2 = 191921.875

Standard deviation = √(191921.875/8 = 154.89

For students without military scholarship,

n = 12

Mean = (820 + 850 + 980 + 1010 + 1020 + 1080 + 1100 + 1120 + 1120 + 1200 + 1220 + 1330)/12

Mean = 1073.83

Summation(x - mean) = (820 - 1073.83)^2 + (850 - 1073.83)^2 + (980 - 1073.83)^2 + (1010 - 1073.83)^2 + (1020 - 1073.83)^2 + (1080 - 1073.83)^2 + (1100 - 1073.83)^2 + (1120 - 1073.83)^2 + (1120 - 1073.83)^2 + (1200 - 1073.83)^2 + (1220 - 1073.83)^2 + (1330 - 1073.83)^2 = 238199.4268

Standard deviation = √(238199.4268/12 = 140.89

We would set up the hypothesis.

The null hypothesis is

H0 : μw = μo H0 : μw - μo = 0

The alternative hypothesis is

Ha : μw ≠ μo Ha : μw - μo ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xw - xo)/√(sw²/nw + so²/no)

From the information given,

xw = 1099.375

xo = 1073.83

sw = 154.89

so = 140.89

nw = 8

no = 12

t = (1099.375 - 1073.83)/√(154.89²/8 + 140.89²/12)

t = 0.37

The formula for determining the degree of freedom is

df = [sw²/nw + so²/no]²/(1/nw - 1)(sw²/nw)² + (1/no - 1)(so²/no)²

df = [154.89²/8 + 140.89²/12]²/(1/8 - 1)(154.89²/8)² + (1/12 - 1)(140.89²/12)² = 21650688.37/1533492.15

df = 14

We would determine the probability value from the t test calculator. It becomes

p value = 0.72

Since the level of significance of 0.05 < the p value of 0.72, we would not reject the null hypothesis.

Therefore, these data do not provide convincing evidence of a difference in SAT scores between students with and without a military scholarship.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

z = (xw - xo) ± z ×√(sw²/nw + so²/no)

For a 95% confidence interval, the z score is 1.96

xw - xo = 1099.375 - 1073.83 = 25.55

z√(sw²/nw + so²/no) = 1.96 × √(154.89²/8 + 140.89²/12) = 1.96 × √2998.86 + 1654.17)

= 133.7

The confidence interval is

25.55 ± 133.7

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Answer:

$4200

Step-by-step explanation:

12% of 35000, because it's between 9,526 and 38,700.

That is .12 * 35000 = $4200

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Amar rakes leaves for his neighbors to earn money. He earned 64 dollars after
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