Answer :
The test statistic is approximately 3.586, and the p-value is approximately 0.0001, leading to the rejection of the null hypothesis, and the conclusion that there is sufficient evidence to reject the claim that the population mean is greater than 51.7.
To calculate the test statistic and p-value for the hypothesis test, you'll need to perform a one-sample t-test. Given the information provided:
- Null Hypothesis (H0): μ = 51.7
- Alternative Hypothesis (H1): μ > 51.7 (One-tailed test)
- Significance level (α): 0.002
- Sample Size (n): 307
- Sample Mean (M): 54.1
- Sample Standard Deviation (SD): 11.6
First, calculate the test statistic (t-score):
[tex]\[ t = \frac{M - \mu}{\frac{SD}{\sqrt{n}}} \][/tex]
[tex]\[ t = \frac{54.1 - 51.7}{\frac{11.6}{\sqrt{307}}} \][/tex]
Calculating this gives us a test statistic of approximately 3.586.
Next, calculate the degrees of freedom (df), which is n - 1 = 306.
Now, you can use the test statistic and the degrees of freedom to find the p-value associated with it. The p-value represents the probability of observing a test statistic as extreme as the one you calculated, assuming the null hypothesis is true. Since this is a one-tailed test where the alternative hypothesis is μ > 51.7, you're interested in the upper tail of the t-distribution.
Using a t-distribution table, calculator, or statistical software, you can find the p-value corresponding to the test statistic. The p-value turns out to be very close to 0.0001 (accurate to four decimal places).
Since the p-value (0.0001) is less than the significance level (0.002), you would reject the null hypothesis.
The test statistic leads to a decision to reject the null hypothesis, and the final conclusion is that there is sufficient evidence to warrant rejection of the claim that the population mean is greater than 51.7.
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The final conclusion is that there is sufficient evidence to warrant rejection of the claim that the population mean is greater than 51.7.
To find the test statistic for the given sample, we can use the formula:
[tex]z = (M - \mu ) / (SD / \sqrt{n} )[/tex], where M is the sample mean, [tex]\mu[/tex] is the hypothesized population mean, SD is the sample standard deviation, and n is the sample size.
Plugging in the values, we get;
[tex]z = (54.1 - 51.7) / (11.6 / \sqrt{307} )[/tex]
[tex]z = 2.091[/tex]
To find the p-value, we compare the test statistic to the critical value of the standard normal distribution for a one-tailed test with a = 0.002. The p-value is the probability of observing a test statistic as extreme or more extreme than the one obtained. We can calculate this using a standard normal distribution table or a statistical software, which gives us a p-value of 0.0187.
Since the p-value is less than the significance level (a = 0.002), we reject the null hypothesis. Therefore, the test statistic leads to a decision to reject the null hypothesis. The final conclusion is that there is sufficient evidence to warrant rejection of the claim that the population mean is greater than 51.7.
Learn more about normal distribution here:
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