A researcher believes that the average size of farms in the U.S. has increased from the 2002 mean of 471 acres. She took a sample of 23 farms in 2011 to test her belief and found a sample mean of 498.8 acres and a sample standard deviation of 46.9 acres. At a 5% levelof significance, test the researcher's claim. What is your conclusion?
2. A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and determines that 310 of these firms have flexible scheduling. At a 1% level of significance, does your test show enough evidence to conclude that a significantly lower percentage of accounting firms offer employees flexible scheduling? What is the p-value for this test?
3. The American Lighting Company developed a new light bulb that it believes will last at least 700 hours on average. A test is to be conducted using a random sample of 100 bulbs, and a 5% level of significance. Assume that the population standard deviation is 15 hours. What are the consequences of making a type II error? What is the probability of making a Type II error if the true population mean is 695 hours?
1. To test the researcher's claim regarding the average size of farms in the U.S., we can perform a hypothesis test.
Step 1: State the hypotheses.
Null hypothesis (H0): The average size of farms in the U.S. has not increased from the 2002 mean of 471 acres.
Alternative hypothesis (Ha): The average size of farms in the U.S. has increased from the 2002 mean of 471 acres.
Step 2: Set the significance level.
Given a 5% level of significance (α = 0.05).
Step 3: Calculate the test statistic.
To conduct this test, we can use the one-sample t-test since the population standard deviation is not known. The formula for the t-test statistic is:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
t = (498.8 - 471) / (46.9 / √23)
Step 4: Determine the critical value.
For a one-tailed test with a 5% level of significance and 22 degrees of freedom (sample size - 1), the critical value can be found in the t-distribution table. In this case, the critical value is approximately 1.717.
Step 5: Make a decision and interpret the results.
If the absolute value of the calculated t-test statistic exceeds the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, the calculated t-test statistic is approximately 3.042. Since 3.042 > 1.717, we reject the null hypothesis. This means there is sufficient evidence to support the researcher's claim that the average size of farms in the U.S. has increased from the 2002 mean of 471 acres.
2. To test whether a significantly lower percentage of accounting firms offer flexible scheduling, we can perform a hypothesis test.
Step 1: State the hypotheses.
Null hypothesis (H0): The percentage of accounting firms offering flexible scheduling is the same as the overall percentage of 79%.
Alternative hypothesis (Ha): The percentage of accounting firms offering flexible scheduling is significantly lower than 79%.
Step 2: Set the significance level.
Given a 1% level of significance (α = 0.01).
Step 3: Calculate the test statistic.
To conduct this test, we can use the z-test since we are comparing proportions. The formula for the z-test statistic is:
z = (sample proportion - population proportion) / √[(population proportion * (1 - population proportion)) / sample size]
z = (310/415 - 0.79) / √[(0.79 * (1 - 0.79)) / 415]
Step 4: Determine the critical value.
For a one-tailed test with a 1% level of significance, the critical value can be found in the standard normal distribution table. In this case, the critical value is approximately -2.326.
Step 5: Make a decision and interpret the results.
If the calculated z-test statistic falls within the critical region (i.e., less than -2.326), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, the calculated z-test statistic is approximately -5.185. Since -5.185 < -2.326, we reject the null hypothesis. This indicates that there is enough evidence to conclude that a significantly lower percentage of accounting firms offer flexible scheduling compared to the overall percentage of 79%.
Regarding the p-value, it represents the probability of observing a test statistic as extreme as or more extreme than the one calculated, assuming the null hypothesis is true. In this case, the p-value is below the 1% level of significance, indicating strong evidence against the null hypothesis. However, the exact p-value cannot be determined without the specific distribution used in the calculation.
3. To calculate the probability of making a Type II error, we need to assume a specific alternative hypothesis. Assuming the alternative hypothesis is that the true population mean is 695 hours, we can proceed with the calculation.
Step 1: Specify the null and alternative hypotheses.
Null hypothesis (H0): The mean lifespan of the new light bulb is equal to 700 hours.
Alternative hypothesis (Ha): The mean lifespan of the new light bulb is less than 700 hours.
Step 2: Set the significance level.
Given a 5% level of significance (α = 0.05).
Step 3: Calculate the critical value.
Since the alternative hypothesis is one-tailed and assumes a decrease in the mean, we can calculate the critical value using the z-distribution. For a one-tailed test with a 5% level of significance, the critical value is approximately -1.645.
Step 4: Calculate the probability of making a Type II error.
To calculate the probability of Type II error (β), we need additional information such as the population mean under the alternative hypothesis, the population standard deviation, and the sample size. However, these values are not provided in the question, making it impossible to determine the exact probability of making a Type II error.
1. To test the researcher's claim, we can perform a hypothesis test. The null hypothesis (H0) is that the average size of farms is still 471 acres, and the alternative hypothesis (Ha) is that the average size has increased. We can use a one-sample t-test for this scenario.
Let's calculate the test statistic (t-value) using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
In this case, the sample mean is 498.8 acres, the hypothesized mean is 471 acres, the sample standard deviation is 46.9 acres, and the sample size is 23 farms.
t = (498.8 - 471) / (46.9 / sqrt(23)) ≈ 2.73
Next, we need to find the critical value for a 5% level of significance. Since this is a two-tailed test, we divide the significance level by 2, resulting in 0.025 for each tail. We can use a t-distribution table or statistical software to find the critical t-value.
With 22 degrees of freedom (sample size - 1), the critical t-value for a 5% level of significance is approximately 2.074 (from the t-distribution table).
Since our calculated t-value of 2.73 is greater than the critical t-value of 2.074, we reject the null hypothesis. There is enough evidence to conclude that the average size of farms has increased from the 2002 mean of 471 acres.
2. To test whether a significantly lower percentage of accounting firms offer flexible scheduling compared to the general figure of 79%, we can perform a hypothesis test. The null hypothesis (H0) is that the percentage of accounting firms offering flexible scheduling is the same as the overall figure of 79%, and the alternative hypothesis (Ha) is that the percentage is lower.
Let's calculate the test statistic (z-value) using the formula:
z = (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size)
In this case, the sample proportion is 310/415 (or 0.747), the hypothesized proportion is 0.79, and the sample size is 415 accounting firms.
z = (0.747 - 0.79) / sqrt((0.79 * (1 - 0.79)) / 415) ≈ -2.33
Next, we need to find the critical value for a 1% level of significance. Since this is a one-tailed test (we want to know if it is significantly lower), we can find the critical z-value corresponding to a 1% upper tail using a standard normal distribution table or statistical software.
The critical z-value for a 1% level of significance is approximately -2.33 (from the standard normal distribution table).
Since our calculated z-value of -2.33 is less than the critical z-value of -2.33, we reject the null hypothesis. There is enough evidence to conclude that a significantly lower percentage of accounting firms offer employees flexible scheduling.
Additionally, to find the p-value for this test, we would compare the calculated z-value to the standard normal distribution. However, since the p-value is not provided, we cannot determine its exact value with the given information.
3. To test if the new light bulb lasts at least 700 hours on average, we can perform a hypothesis test. The null hypothesis (H0) is that the mean duration of the light bulbs is 700 hours or more, and the alternative hypothesis (Ha) is that the mean duration is less than 700 hours.
We can calculate the test statistic (z-value) using the formula:
z = (sample mean - hypothesized mean) / (population standard deviation / sqrt(sample size))
In this case, the hypothesized mean is 700 hours, the population standard deviation is 15 hours, and the sample size is 100 bulbs.
z = (sample mean - 700) / (15 / sqrt(100)) = (sample mean - 700) / 1.5
To find the critical value for a 5% level of significance, we can use a standard normal distribution table or statistical software. For a one-tailed test where we are interested in whether the mean is less than 700 hours, the critical z-value for a 5% level of significance is approximately -1.645.
Now, to calculate the probability of making a Type II error (β) if the true population mean is 695 hours, we need to determine the probability of failing to reject the null hypothesis (H0) when H0 is false (i.e., the mean is actually less than 700 hours).
β = P(Failed to reject H0 | H0 is false)
This probability depends on the specific alternative hypothesis (Ha) and the difference between the true population mean and the hypothesized mean. Without knowing the alternative hypothesis, we cannot provide a specific value for β in this case.