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.
Gagyaq
the answer would be b.
To test the hypotheses H0: m = 70, Ha: m > 70, we can use a one-sample t-test. The test statistic is calculated as:
t = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ = sample mean
μ = population mean under the null hypothesis
s = sample standard deviation
n = sample size
Given that:
x̄ = (79 + 76 + 72 + 73) / 4 = 75
μ = 70
s = 2.4
n = 4
Now, we can calculate the test statistic:
t = (75 - 70) / (2.4 / sqrt(4))
= 5 / (2.4 / 2)
= 5 / 1
= 5
Next, we need to determine whether to reject the null hypothesis based on the test statistic and the desired significance level.
a. If the significance level is 0.10, the critical value for a one-tailed test is 1.323 (from t-distribution with n-1 = 4-1 = 3 degrees of freedom). Since 5 > 1.323, we would reject H0 at level 0.10.
b. If the significance level is 0.05, the critical value for a one-tailed test is 2.131 (from t-distribution with n-1 = 4-1 = 3 degrees of freedom). Since 5 > 2.131, we would reject H0 at level 0.05.
c. If the significance level is 0.01, the critical value for a one-tailed test is 3.182 (from t-distribution with n-1 = 4-1 = 3 degrees of freedom). Since 5 > 3.182, we would reject H0 at level 0.01.
Therefore, the correct answer is:
c. I would reject H0 at level 0.01.
To test the hypotheses H0: m = 70 vs Ha: m > 70, we can perform a one-sample t-test using the given data.
Step 1: State the hypotheses
H0: m = 70 (null hypothesis)
Ha: m > 70 (alternative hypothesis)
Step 2: Set the significance level
This problem doesn't explicitly provide the significance level (α) to use for the test. Let's assume α = 0.05 for this explanation.
Step 3: Compute the test statistic
To perform the one-sample t-test, we first need to calculate the sample mean (x̄) and the sample standard deviation (s). From the given data, the sample mean is (79+76+72+73)/4 = 75. The sample standard deviation can be calculated using the formula: s = √[Σ(xi - x̄)²/(n-1)]. Plugging in the values, we get s = 2.58 (rounding to two decimal places).
Next, we calculate the test statistic using the formula: t = (x̄ - μ) / (s / √n), where μ is the hypothesized mean. In this case, μ = 70 and n = 4. Plugging in the values, we get t = (75 - 70) / (2.58 / √4) = 3.875 (rounding to three decimal places).
Step 4: Determine the p-value
To determine the p-value associated with the test statistic, we need to find the area under the t-distribution curve to the right of our test statistic. Since the sample size is small (n < 30), we use the t-distribution to calculate the p-value. Using a t-table or statistical software, we find that the p-value is approximately 0.011.
Step 5: Make a decision
Comparing the p-value (0.011) to the significance level (0.05), we can make a decision. Since the p-value is less than the significance level, we reject the null hypothesis.
Answer:
c. I would reject H0 at level 0.01.
Note: The correct answer may vary depending on the provided significance level in the problem or whether the p-value is provided.