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A simple random sample of 20 pages from a dictionary is obtained. The numbers of words defined on those pages are found, with the results [tex]$n=20, \bar{x}=59.3$[/tex] words, [tex]$s=15.9$[/tex] words. Given that this dictionary has 1481 pages with defined words, the claim that there are more than 70,000 defined words is equivalent to the claim that the mean number of words per page is greater than 47.3 words. Use a 0.10 significance level to test the claim that the mean number of words per page is greater than 47.3 words. What does the result suggest about the claim that there are more than 70,000 defined words? Identify the null and alternative hypotheses, test statistic, [tex]$P$[/tex]-value, and state the final conclusion that addresses the original claim. Assume that the population is normally distributed.

A. [tex]$H_0: \mu = 47.3$[/tex] words

B. [tex]$H_1: \mu \ \textgreater \ 47.3$[/tex] words

Determine the test statistic.
(Round to two decimal places as needed.)

Sagot :

GinnyAnswer
Sure, let's work through this problem step-by-step:

### Step 1: Identify the Hypotheses
We need to set up our null and alternative hypotheses based on the claim that the mean number of words per page in the dictionary is greater than 47.3 words.

- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean number of words per page is 47.3 words.
[tex]\[ H_0: \mu = 47.3 \text{ words} \][/tex]

- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean number of words per page is greater than 47.3 words.
[tex]\[ H_1: \mu > 47.3 \text{ words} \][/tex]

### Step 2: Gather Sample Statistics
From the problem, we know:
- Sample size ([tex]\(n\)[/tex]) = 20
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 59.3 words
- Sample standard deviation ([tex]\(s\)[/tex]) = 15.9 words

### Step 3: Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is given as 0.10.

### Step 4: Test Statistic Calculation
The test statistic for a one-sample t-test can be calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]

Given:
- [tex]\(\bar{x} = 59.3\)[/tex]
- [tex]\(\mu = 47.3\)[/tex]
- [tex]\(s = 15.9\)[/tex]
- [tex]\(n = 20\)[/tex]

Plugging in these values, we get:
[tex]\[ t = \frac{59.3 - 47.3}{15.9 / \sqrt{20}} \][/tex]
[tex]\[ t \approx 3.38 \][/tex]

### Step 5: Degrees of Freedom
The degrees of freedom (df) for this test is:
[tex]\[ df = n - 1 = 20 - 1 = 19 \][/tex]

### Step 6: Calculate the P-value
The P-value can be found using the cumulative distribution function (CDF) for the t-distribution. For a one-tailed test, we want to find the area to the right of the calculated t value.

Given that the test statistic [tex]\(t \approx 3.38\)[/tex] and the degrees of freedom [tex]\(df = 19\)[/tex], we find:
[tex]\[ P \text{-value} \approx 0.0016 \][/tex]

### Step 7: Make a Decision
We compare the P-value to the significance level ([tex]\(\alpha = 0.10\)[/tex]).

Since [tex]\(0.0016 < 0.10\)[/tex], we reject the null hypothesis [tex]\(H_0\)[/tex].

### Conclusion
Since we have rejected the null hypothesis, we have enough evidence to support the claim that the mean number of words per page is greater than 47.3 words.

#### Result:
- Test Statistic: [tex]\(t \approx 3.38\)[/tex]
- P-value: [tex]\(\approx 0.0016\)[/tex]
- Conclusion: Reject the null hypothesis. There is sufficient evidence to support the claim that the mean number of words per page is greater than 47.3 words, which suggests that there could indeed be more than 70,000 defined words in the dictionary.
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