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myrzilka [38]
3 years ago
7

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 5.9 parts

/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 24 samples is 5.6 ppm with a standard deviation of 1.0. Assume the population is normally distributed. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to three decimal places.
Mathematics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

t = -0.061

Step-by-step explanation:

We are given;

Population mean; μ = 5.9

Sample mean; x¯ = 5.6

Sample size; n = 24

standard deviation; σ = 1

Significance level = 0.02

Formula for the test statistic since sample size is less than 30 is;

t = (x¯ - μ)/(σ/√n)

t = (5.6 - 5.9)/(1/√24)

t = -0.3/(1/√24)

t = -0.061

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shtirl [24]

Answer:

The vertex of the given equation is    which shows that the value of the market after about 5 years is about 187,253.01

Functions and values

Given the overall value of a home can be modeled by

V(x) = 415x^2 – 4600x + 200000

Write in vertex form to have:

V(x) = 415x^2 – 4600x + 200000

For the given equation, the vertex occur at

Hence the vertex of the given equation is    which shows that the value of the market after about 5 years is about 187,253.01

4 0
2 years ago
VIOLI me eg 20(20-x) = 300 What is the value of x? Round to the units place.​
vivado [14]

Answer:

x=5

Step-by-step explanation:

20(20-x) = 300

Divide each side by 20

20/20(20-x) = 300/20

20-x = 15

Subtract 20 from each side

20-x-20 = 15-20

-x = -5

Multiply each side by -1

x = 5

8 0
2 years ago
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On Monday, Ken spent half of his money on a new game. The next day he earned $12 mowing lawns. He now has $32. How much money di
saveliy_v [14]

Answer:

40.00000000000000000

3 0
3 years ago
I don't get it.... I checked my book for the answers and it shows usless info
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Answer:

x= 22.6

Step-by-step explanation:

This little red square means that is a right angle (90 degrees)  

x is the complementary of 67.4  (Both add to 90)

90-67.4 = 22.6

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3 years ago
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A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classe
stealth61 [152]

Answer:

t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890  

The p value for this case would be:

p_v =P(t_{18}>0.890)=0.193  

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B

Step-by-step explanation:

Information given

\bar X_{1}=4 represent the mean for sample A

\bar X_{2}=3.5 represent the mean for sample B

s_{1}=1.5 represent the sample standard deviation for A

s_{2}=1 represent the sample standard deviation for B  

n_{1}=11 sample size for the group A

n_{2}=9 sample size for the group B  

\alpha=0.1 Significance level provided

t would represent the statistic

Hypothesis to test

We want to verify if the student who graduates from college A has taken more math classes, on the average, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2} \leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

The statistic is given by:

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=11+9-2=18  

Replacing we got:

t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890  

The p value for this case would be:

p_v =P(t_{18}>0.890)=0.193  

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B

5 0
3 years ago
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