Sample Of N=16 From A Normal Population

If, in a sample of n=16 selected from a normal population, Xbar = 56 and S = 12, what is the value of t stat if you are testing the null hypothesis Ho: mu = 50?

In this statistical problem, we are asked to calculate the t-statistic for a small sample drawn from a normal population, with a specific null hypothesis about the population mean. The given data includes a sample size (n = 16), sample mean (X̄ = 56), and sample standard deviation (S = 12). The null hypothesis (H₀) states that the population mean (μ) equals 50. To determine whether there's evidence to reject this null hypothesis, we perform a t-test.

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The t-test is a vital inferential statistic used when the sample size is small, and the population standard deviation is unknown. It helps determine whether there is a significant difference between the sample mean and the hypothesized population mean under the null hypothesis. In this case, the parameters are as follows: n = 16, X̄ = 56, S = 12, and H₀: μ = 50.

The formula for the t-statistic in a one-sample t-test is:

t = (X̄ - μ₀) / (S / √n)

where X̄ is the sample mean, μ₀ is the hypothesized population mean, S is the sample standard deviation, and n is the sample size. Plugging the values into the formula:

t = (56 - 50) / (12 / √16)

Calculating the denominator:

12 / √16 = 12 / 4 = 3

Therefore, the t-statistic is:

t = 6 / 3 = 2

This t-value of 2 indicates how many standard errors the sample mean is away from the hypothesized population mean under the null hypothesis. It is essential to compare this t-value against critical t-values from the t-distribution table to determine significance.

Degrees of Freedom

The degrees of freedom for this test are calculated as n - 1, which in this case is:

df = 16 - 1 = 15

This value influences the shape of the t-distribution and the critical value used for hypothesis testing.

Critical Values at α = 0.05

For a two-tailed test at a significance level of 0.05 and 15 degrees of freedom, the critical t-values are approximately ±2.131 (from standard t-distribution tables). Comparing our calculated t-value of 2 to these critical values, we see that 2 is less than 2.131, suggesting that the result may not be statistically significant at the 0.05 level, but close.

P-value Calculation

The P-value corresponds to the probability of observing a t-value as extreme or more extreme than the calculated value under the null hypothesis. Using a t-distribution calculator or software for t = 2 with 15 degrees of freedom, the two-tailed P-value is approximately 0.062.

Since the P-value (≈0.062) exceeds the significance level of 0.05, we fail to reject the null hypothesis at the 5% significance level. This suggests insufficient evidence to conclude that the population mean differs significantly from 50 based on this sample.

Summary

  • Calculated t-statistic: 2
  • Degrees of freedom: 15
  • Critical t-value (α=0.05, two-tailed): ±2.131
  • P-value: approximately 0.062
  • Decision: Fail to reject H₀; there's not enough evidence to suggest a significant difference.

Conclusion

In summary, based on the sample data, the computed t-statistic is 2, which is just below the critical value for a two-tailed test at the 0.05 significance level, and the P-value is slightly above 0.05. Consequently, the evidence is insufficient to reject the null hypothesis that the population mean is 50, indicating that the observed difference could plausibly be due to sampling variability.

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