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3 years ago

Your best friend, who is on spring break in Europe, e-mailed you that the temperature in Paris today is . Use the formula F

Mathematics

1 answer:

kifflom [539]3 years ago

7 0

Answer:

<h2> Assuming your friend said the temperature over there was 68 degrees Fahrenheit in degrees C it would be 20 °C</h2>

Step-by-step explanation:

The question is not complete, as the information, your best friend in Europe about the temperature over there is missing, nevertheless, we can proceed with this problem by plugging in an assumed temperature.

The expression for conversion from degrees F to degrees C is given as

T(°C) = (T(°F) - 32) × 5/9

Now say temperature your friend e-mailed you is 68 degrees Fahrenheit (you can as well plug in your own value)

T(°C) = (68°F - 32) × 5/9

T(°C) = (36) × 5/9

T(°C) = 180/9 = 20 °C

T(°C) = 20 °C

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$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

-The sample is too small to make judgments about skewness or symmetry.

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003 d2=1.337-1.322=0.015

d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

And for the sample deviation we can use the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

$\bar X_{1}=1.173$ represent the mean for the operator 1

$\bar X_{2}=1.174$ represent the mean for the operator 2

$s_{1}=0.1506$ represent the sample standard deviation for the operator 1

$s_{2}=0.1495$ represent the sample standard deviation for the operator 2

$n_{1}=8$ sample size for the operator 1

$n_{2}=8$ sample size for the operator 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) like this:

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case: