Answer:
Option C
Step-by-step explanation:
we know that
The area of a circle is equal to
we have
substitute
Remember that
degrees subtends the complete area of the circle
so
by proportion
Find the area of the blue sector for an angle of degrees
Which of the following formulas could be used to find the perimeter, P, of a regular hexagon? brainleist
1 answer:
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Answer:
(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.
(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.
(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.
(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.
Step-by-step explanation:
To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.
The significance level for the test is <em>α</em> = 0.05.
The decision rule is:
If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.
(a)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>
The sample size is, <em>n</em> = 11.
The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.
The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₁₀ > 1.91) = 0.043.
The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
(b)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>
The sample size is, <em>n</em> = 17.
The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.
The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.
The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
(c)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>
The sample size is, <em>n</em> = 7.
The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.
The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.
The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.
The null hypothesis is not rejected at 5% level of significance.
(d)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>
The sample size is, <em>n</em> = 28.
The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.
The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₂₇ > 2.12) = 0.022.
The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
