Mathematics, 17.06.2020 22:57 jose477

A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that your team scored at an average of less than 110 points in the years 2013-2015. Test this claim at a 1% level of significance. For this test, assume that the population standard deviation for relative skill level is unknown.

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Answer from: najashohatee1234

There is not enough evidence to support the claim that your team scored an average significantly less than 110 points.

Step-by-step explanation:

The question is incomplete:

There is no data from the sample.

We will use the sample [105107117106110 ] as an example to solve the question.

The mean of the sample is:

$M=\dfrac{1}{5}\sum_{i=1}^{5}(105+107+117+106+110)\\\\\\ M=\dfrac{545}{5}=109$

The standard deviation of the sample is:

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(105-(109))^2+(107-(109))^2+(117-(109))^2+(106-(109))^2+(110-(109))^2]}\\\\\\ s=\sqrt{\dfrac{1}{4}\cdot [(16)+(4)+(64)+(9)+(1)]}\\\\\\ s=\sqrt{\dfrac{94}{4}}=\sqrt{23.5}\\\\\\s=4.848$

This is a hypothesis test for the population mean.

The claim is that your team scored an average significantly less than 110 points.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu< 110$

The significance level is 0.01.

The sample has a size n=5.

The sample mean is M=109.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.848.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.848}{\sqrt{5}}=2.17$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{109-110}{2.17}=\dfrac{-1}{2.17}=-0.46$

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-0.46, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.334) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that your team scored an average significantly less than 110 points.

Answer from: Quest

Magic man is here, but unfortunately he is broke won’t you him out

Answer from: Quest

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

The graph below shows the value of a $100 deposited into three different accounts over a period of 2

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