Respuesta :
Answer:
(a) p-value = 0.043. Null hypothesis is rejected.
(b) p-value = 0.001. Null hypothesis is rejected.
(c) p-value = 0.444. Null hypothesis is not rejected.
(d) p-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 t-test for single mean is used.
The significance level for the test is α = 0.05.
The decision rule is:
If the p - value is less than the significance level then the null hypothesis will be rejected. And if the p-value is more than the value of α then the null hypothesis will not be rejected.
(a)
The alternate hypothesis is:
Hₐ: μ > μ₀
The sample size is, n = 11.
The test statistic value is, t = 1.91 ≈ 1.90.
The degrees of freedom is, (n - 1) = 11 - 1 = 10.
Use a t-table t compute the p-value.
For the test statistic value of 1.90 and degrees of freedom 10 the p-value is:
The p-value is:
P (t₁₀ > 1.91) = 0.043.
The p-value = 0.043 < α = 0.05.
The null hypothesis is rejected at 5% level of significance.
(b)
The alternate hypothesis is:
Hₐ: μ < μ₀
The sample size is, n = 17.
The test statistic value is, t = -3.45 ≈ 3.50.
The degrees of freedom is, (n - 1) = 17 - 1 = 16.
Use a t-table t compute the p-value.
For the test statistic value of -3.50 and degrees of freedom 16 the p-value is:
The p-value is:
P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.
The p-value = 0.001 < α = 0.05.
The null hypothesis is rejected at 5% level of significance.
(c)
The alternate hypothesis is:
Hₐ: μ ≠ μ₀
The sample size is, n = 7.
The test statistic value is, t = 0.83 ≈ 0.82.
The degrees of freedom is, (n - 1) = 7 - 1 = 6.
Use a t-table t compute the p-value.
For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:
The p-value is:
P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.
The p-value = 0.444 > α = 0.05.
The null hypothesis is not rejected at 5% level of significance.
(d)
The alternate hypothesis is:
Hₐ: μ > μ₀
The sample size is, n = 28.
The test statistic value is, t = 2.13 ≈ 2.12.
The degrees of freedom is, (n - 1) = 28 - 1 = 27.
Use a t-table t compute the p-value.
For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:
The p-value is:
P (t₂₇ > 2.12) = 0.022.
The p-value = 0.444 > α = 0.05.
The null hypothesis is rejected at 5% level of significance.
Using the t-distribution, it is found that:
a) The p-value is of 0.0426, which is less than 0.05, thus the null hypothesis is rejected.
b) The p-value is of 0.0016, which is less than 0.05, thus the null hypothesis is rejected.
c) The p-value is of 0.4383, which is more than 0.05, thus the null hypothesis is not rejected.
d) The p-value is of 0.0195, which is less than 0.05, thus the null hypothesis is rejected.
Item a:
- We have a right-tailed test, as it is being tested if the mean is more than a value.
The p-value is found using a t-distribution calculator, with 11 - 1 = 10 df and t = 1.91, hence it is of 0.0426.
The p-value is of 0.0426, which is less than 0.05, thus the null hypothesis is rejected.
Item b:
- We have a left-tailed test, as it is being tested if the mean is less than a value.
The p-value is found using a t-distribution calculator, with 17 - 1 = 16 df and t = -3.45, hence it is of 0.0016.
The p-value is of 0.0016, which is less than 0.05, thus the null hypothesis is rejected.
Item c:
- We have a two-tailed test, as it is being tested if the mean is different than a value.
The p-value is found using a t-distribution calculator, with 7 - 1 = 6 df and t = 0.83, hence it is of 0.4383.
The p-value is of 0.4383, which is more than 0.05, thus the null hypothesis is not rejected.
Item d:
- We have a right-tailed test, as it is being tested if the mean is more than a value.
The p-value is found using a t-distribution calculator, with 28 - 1 = 27 df and t = 1.91, hence it is of 0.0195.
The p-value is of 0.0195, which is less than 0.05, thus the null hypothesis is rejected.
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