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3 years ago

Help me with this question please

Mathematics

1 answer:

Fed [463]3 years ago

8 0

Answer:

--->(f.g)(x)=8(x-1)^2+7(x-1)-9

=8(x^2-2x+1)+7x-7-9

=8x^2-9x-8

--->(f.g)(-2)=8(-2)^2-9(-2)-8=42

✌️:)

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The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

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Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

$H_0: \mu_d=0\\\\H_a: \mu_d>0$

We have n=8 pairs of data. We calculate the difference as:

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$d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]$

The sample mean and standard deviation are:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\$

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729$

The degrees of freedom for this sample size are:

$df=n-1=8-1=7$

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>1.729)=0.0637$

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

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