Suppose that: x = 197.1, μ = 191, α = 5%, δ = 25.6, η = 100. Test the claims that, population mean is more than 191.
To test the claim that the population mean is more than 191, we will use a one-sample t-test.
Given:
x = 197.1 (sample mean)
μ = 191 (population mean)
α = 0.05 (significance level)
δ = 25.6 (standard deviation)
η = 100 (sample size)
Hypotheses:
Null Hypothesis (H0): μ ≤ 191
Alternative Hypothesis (Ha): μ > 191
Test Statistic:
t = (x - μ) / (δ / sqrt(η))
t = (197.1 - 191) / (25.6 / sqrt(100))
t = 6.1 / (25.6 / 10)
t = 6.1 / 2.56
t = 2.39
Critical Value:
Since α = 0.05 is the significance level, the critical value will be obtained from the t-distribution table with 99 degrees of freedom (100 - 1).
The critical value for a right-tailed test with α = 0.05 and 99 degrees of freedom is approximately 1.660
Decision:
If the test statistic (t) is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Calculation:
2.39 > 1.660
Since 2.39 is greater than 1.660, we reject the null hypothesis.
Conclusion:
Based on the sample data, with a 5% significance level, we have enough evidence to conclude that the population mean is more than 191.