2. The time (in number of days) until maturity of a certain variety of hot pepper is normally distributed, with mean m and standard deviation s = 2.4. This variety is advertised as taking 70 days to mature. I wish to test the hypotheses H0: m = 70, Ha: m > 70, so I select a simple random sample of four plants of this variety and measure the time until maturity. The four times, in days, are
79 76 72 73.
Based on these data
a. I would reject H0 at level 0.10 but not at level 0.05.
b. I would reject H0 at level 0.05 but not at level 0.01.
c. I would reject H0 at level 0.01.
d. None of the above.
To test the hypotheses H0: m = 70, Ha: m > 70, we can use a one-sample t-test.
First, let's calculate the sample mean and standard deviation for the given data:
Sample mean (x̄) = (79 + 76 + 72 + 73) / 4 = 75
Sample standard deviation (s) = √[((79-75)² + (76-75)² + (72-75)² + (73-75)²) / (4-1)] = √[(16 + 1 + 9 + 4) / 3] = √[30/3] = √10 ≈ 3.16
Next, we need to calculate the test statistic, which is the t-value. The formula for the t-value in a one-sample t-test is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized mean (70 in this case), s is the sample standard deviation, and n is the sample size (4 in this case).
t = (75 - 70) / (3.16 / √4) = 5 / (3.16 / 2) ≈ 5 / 1.58 ≈ 3.16
Finally, we need to determine the critical t-value for the given significance levels:
For a level of 0.10 with a one-tailed test and degrees of freedom (df) = n - 1 = 4 - 1 = 3, the critical t-value is approximately 1.638.
For a level of 0.05 with a one-tailed test and df = 3, the critical t-value is approximately 2.353.
For a level of 0.01 with a one-tailed test and df = 3, the critical t-value is approximately 3.182.
Comparing the calculated t-value (3.16) with the critical t-values, we can determine:
a. The calculated t-value of 3.16 exceeds the critical t-value of 1.638 at the 0.10 level, so we would reject H0 at level 0.10.
b. The calculated t-value of 3.16 also exceeds the critical t-value of 2.353 at the 0.05 level, so we would reject H0 at level 0.05.
c. The calculated t-value of 3.16 exceeds the critical t-value of 3.182 at the 0.01 level, so we would reject H0 at level 0.01.
Therefore, the correct answer is:
c. I would reject H0 at level 0.01.