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charle [14.2K]
3 years ago
13

A research scientist measures the outside temperature, in degrees Fahrenheit, at various times throughout a 24-hour period and r

ecords her findings in the graph below. In this real-world function, the hours represent the input values, and the temperature represents the output values. Use the graph to identify the range of the function. Write your answers using set notation. Then write a sentence to explain what the range represents in this function.
Mathematics
1 answer:
Cloud [144]3 years ago
4 0

Answer:

Step-by-step explanation:

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Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found
Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let X_{ij} denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

Xbar_{i} =\frac{1}{J_{i}} summation(X_{ij}) where J=1 to J_{i} and  J_{i}  is ith sample size

s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} -   Xbar_{i})^{2}

The sample sizes are J₁ =12   J₂=10   J₃=18   J₄=9

the total number in all samples combined is  49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar=\frac{1}{49} (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ J_{i} (Xbar_{i} -Xbar)^{2}  from i=1 to 49

SSTr= 20.9910

SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}

SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

F=\frac{MSTr}{MSE}

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an F_{3,45} distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0
3 years ago
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