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3 years ago

At 6 a.m the temperature was -8.At noon the temperature was 3.What was the change of temperature between 6 a.m and noon

Mathematics

2 answers:

Damm [24]3 years ago

4 0

11 degrees
3- -8 = 3 + 8 = 11

Oduvanchick [21]3 years ago

4 0

If it was -8 (I'm assuming you're talking Celsius right?) then it'd have to go up 8 to get to 0, after that you would add three. The answer should be, it went up 11 degrees. I think lol

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Zanzabum

Answer:

a) Null hypothesis: $\mu \leq 40000$

Alternative hypothesis: $\mu > 40000$

b) $t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$

c) The first step is calculate the degrees of freedom, on this case:

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d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance. So we can conclude that the true mean is higher than 40000 at the significance level assumed.

Step-by-step explanation:

Data given and notation

$\bar X=41182$ represent the sample mean

$s=19990$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =40000$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is greater than 40000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 40000$

Alternative hypothesis: $\mu > 40000$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$

Part c: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since the sample size is large enough we cna use the z distribution as an approximation for the statsitic on this case.

Since is a one right tailed test the p value would be:

$p_v =P(t_{(1699)}>2.438)=0.0074$

Part d: Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance. So we can conclude that the true mean is higher than 40000 at the significance level assumed.

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Answer:

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Step-by-step explanation:

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