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3 years ago

The 9th term of an arithmetic series is 5/2

Mathematics

1 answer:

Verdich [7]3 years ago

8 0

Answer:

- 8050

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a + (n - 1)d

where a is the first term and d the common difference.

We require to find both a and d

Given the 9 th term is 2.5 , then

a + 8d = 2.5 → (1)

Given the sum of the second and fifth term is 27, then

a + d + a + 4d = 27, that is

2a + 5d = 27 → (2)

Multiply (1) by - 2 and add to (2) to eliminate a

- 2a - 16d = - 5 → (3)

Add (2) and (3) term by term

- 11d = 22 ( divide both sides by - 11 )

d = - 2

Substitute d = - 2 into (1) and solve for a

a - 16 = 2.5 ( add 16 to both sides )

a = 18.5

The sum to n terms of an arithmetic sequence is

$S_{n}$ = $\frac{n}{2}$ [ 2a + (n - 1)d ], thus

$S_{100}$ = 50 [ (2 × 18.5) + (99 × - 2) ] = 50(37 - 198) = 50(- 161) = - 8050

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Alternative hypothesis: $\mu > 40000$

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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)

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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:

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Part d: Conclusion

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