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raketka [301]
3 years ago
7

The temperature in degrees Celsius can be expressed by the function C(f)=59(f−32)C(f)=59(f−32) where f is the temperature in deg

rees Fahrenheit. Find the temperature in degrees Celsius (to the nearest degree) if it is 55 °f.
Mathematics
1 answer:
Fiesta28 [93]3 years ago
4 0
\dfrac{5}{9}\cdot (55-32)=\dfrac{5}{9}\cdot 23=\dfrac{115}{9}\approx 13^\circ C
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Answer:

F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_2 -1 =16-1=15 and for the denominator we have n_1 -1 =16-1=15 and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

p_v =2*P(F_{15,15}>2.2)=0.138

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

Step-by-step explanation:

Information given

n_1 = 16 represent the sampe size 1

n_2 =16 represent the sample 2

s^2_1 = 2.5 represent the sample deviation for 1

s^2_2 = 5.5 represent the sample variance for 2

\alpha=0.10 represent the significance level provided

The statistic is given by:

F=\frac{s^2_2}{s^2_1}

Hypothesis to test

We want to test if the variations in terms of the variance are equal, so the system of hypothesis are:

H0: \sigma^2_1 = \sigma^2_2

H1: \sigma^2_1 \neq \sigma^2_2

The statistic is given by:

F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_2 -1 =16-1=15 and for the denominator we have n_1 -1 =16-1=15 and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

p_v =2*P(F_{15,15}>2.2)=0.138

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

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