Helen had 5/6 can of orange juice. She spilled 2/6 of a can. How much juice did Helen have left?
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.
