Suppose that the average grade for all people who take this course in statistics is 83 with a standard deviation of 3.5. If your particular class has an average of 89, can you conclude that your class is better than average at the 90 percent confidence level? Sample size 21.
Identify one or two-tailed
States the Null and Alternative Hypothesis
States the Level of Significance
Calculates Test Statistic
Formulates Decision Rule
Makes a Conclusion
Null hypothesis:
Ho: µ = 83 -->meaning: population mean is equal to 83
Ha: µ does not equal 83 -->meaning: population does not equal 83
Since the problem is using confidence intervals, remember that 5% (or .05) will be below the interval and 5% (or .05) will be above the interval. This will be a two-tailed test.
Using the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (89 - 83)/(3.5/√21)
Finish the calculation.
Determine whether or not to reject the null. If you reject the null, you will conclude a difference.
I hope this will help get you started.
To determine whether your class is better than average at the 90 percent confidence level, we can follow these steps:
1. Identify one or two-tailed:
Since we are comparing whether your class is better than the average, this is a one-tailed test. We are only interested in the performance being higher, not lower.
2. State the Null and Alternative Hypothesis:
Null Hypothesis (H0): The average grade of your class is not significantly different from the average grade for all people taking the course (µ = 83).
Alternative Hypothesis (Ha): The average grade of your class is significantly higher than the average grade for all people taking the course (µ > 83).
3. State the Level of Significance:
The level of significance is the probability of rejecting the null hypothesis when it is true. In this case, the level of significance is 0.10 or 10% (90 percent confidence level).
4. Calculate the Test Statistic:
To calculate the test statistic (z-score), we can use the formula:
z = (X̄ - µ) / (σ / √n)
where X̄ is the sample mean (89), µ is the population mean (83), σ is the population standard deviation (3.5), and n is the sample size (21).
Substituting the values: z = (89 - 83) / (3.5 / √21)
Calculate the result of z.
5. Formulate the Decision Rule:
The decision rule is based on the level of significance and the assumption of a normal distribution. For a one-tailed test with a level of significance of 0.10, we would compare the test statistic (z) against the critical value for α = 0.10.
Lookup the critical value for a one-tailed test (90% confidence level) using a standard normal distribution table or a z-table. Compare the calculated test statistic (z) to this critical value.
6. Make a Conclusion:
If the calculated test statistic (z) is greater than the critical value, we reject the null hypothesis. This means that there is evidence to support the alternative hypothesis, suggesting that your class is better than average at the 90 percent confidence level. If the calculated test statistic is less than the critical value, we fail to reject the null hypothesis and do not have sufficient evidence to conclude that your class is better than average.