Asked by Shay
A simple random sample of 10 pages from a dictionary is obtained and the number of defined words on each page is recorded. The sample mean is found to be 53.5 definitions per page and a standard deviation of 15.7 definitions per page. To support a claim that there are more than 70,000 defined words is the same as the claim that the mean number of defined words on a page is greater than 48.0. Use a 0.05 significance level to test this claim.
A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.
B) Fail to reject the null hypothesis; there is sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.
C) Reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is less than 48.0.
D) Reject the null hypothesis; there is sufficient evidence to support the claim that the mean number of defined words on a page is less than 48.0.
A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.
B) Fail to reject the null hypothesis; there is sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.
C) Reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is less than 48.0.
D) Reject the null hypothesis; there is sufficient evidence to support the claim that the mean number of defined words on a page is less than 48.0.
Answers
Answered by
GPT-4o mini
To test the claim that the mean number of defined words on a page is greater than 48.0, we will perform a hypothesis test using the following steps:
1. **State the hypotheses:**
- Null Hypothesis (H0): \(\mu \leq 48.0\) (the mean number of defined words is less than or equal to 48.0)
- Alternative Hypothesis (H1): \(\mu > 48.0\) (the mean number of defined words is greater than 48.0)
2. **Determine the significance level:**
- The significance level is \(\alpha = 0.05\).
3. **Calculate the test statistic:**
We will use a one-sample t-test for the sample mean. The test statistic is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
where:
- \(\bar{x} = 53.5\) (sample mean)
- \(\mu_0 = 48.0\) (hypothesized population mean)
- \(s = 15.7\) (sample standard deviation)
- \(n = 10\) (sample size)
Let's calculate the test statistic:
\[
t = \frac{53.5 - 48.0}{15.7 / \sqrt{10}} = \frac{5.5}{15.7 / 3.1623} \approx \frac{5.5}{4.964} \approx 1.11
\]
4. **Determine the degrees of freedom:**
The degrees of freedom (df) for this test is \(n - 1 = 10 - 1 = 9\).
5. **Find the critical value:**
We will look up the critical value for a one-tailed t-test at \(\alpha = 0.05\) with 9 degrees of freedom. The critical value (from a t-distribution table) is approximately \(t_{0.05, 9} \approx 1.833\).
6. **Make the decision:**
- If the calculated \(t\)-statistic is greater than the critical value, we reject the null hypothesis.
- If it is less than or equal to the critical value, we fail to reject the null hypothesis.
Our calculated \(t\)-statistic of \(1.11\) is less than the critical value of \(1.833\).
7. **Conclusion:**
Since \(1.11 < 1.833\), we fail to reject the null hypothesis.
Therefore, the correct choice is:
**A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.**
1. **State the hypotheses:**
- Null Hypothesis (H0): \(\mu \leq 48.0\) (the mean number of defined words is less than or equal to 48.0)
- Alternative Hypothesis (H1): \(\mu > 48.0\) (the mean number of defined words is greater than 48.0)
2. **Determine the significance level:**
- The significance level is \(\alpha = 0.05\).
3. **Calculate the test statistic:**
We will use a one-sample t-test for the sample mean. The test statistic is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
where:
- \(\bar{x} = 53.5\) (sample mean)
- \(\mu_0 = 48.0\) (hypothesized population mean)
- \(s = 15.7\) (sample standard deviation)
- \(n = 10\) (sample size)
Let's calculate the test statistic:
\[
t = \frac{53.5 - 48.0}{15.7 / \sqrt{10}} = \frac{5.5}{15.7 / 3.1623} \approx \frac{5.5}{4.964} \approx 1.11
\]
4. **Determine the degrees of freedom:**
The degrees of freedom (df) for this test is \(n - 1 = 10 - 1 = 9\).
5. **Find the critical value:**
We will look up the critical value for a one-tailed t-test at \(\alpha = 0.05\) with 9 degrees of freedom. The critical value (from a t-distribution table) is approximately \(t_{0.05, 9} \approx 1.833\).
6. **Make the decision:**
- If the calculated \(t\)-statistic is greater than the critical value, we reject the null hypothesis.
- If it is less than or equal to the critical value, we fail to reject the null hypothesis.
Our calculated \(t\)-statistic of \(1.11\) is less than the critical value of \(1.833\).
7. **Conclusion:**
Since \(1.11 < 1.833\), we fail to reject the null hypothesis.
Therefore, the correct choice is:
**A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.**