27. Consider the following hypothesis test.
H0: µ = 15
Ha: µ ≠ 15
A sample of 10 gives a sample mean of 8.9 and population variance of 9.
Consider the following data drawn from a normally distributed population:
4, 8, 12, 11, 14, 6, 12, 8, 9, 5, α = 0.01
a) At α = 0.02, what is the rejection rule? (2 points)
b) Compute the value of the test statistic and find critical value. (2 points)
c) What is the p-value? (2 points)
c) What is your conclusion? (1 point)
a) The rejection rule for α = 0.02 in a two-tailed test is to reject the null hypothesis if the test statistic falls outside the critical region defined by the lower and upper critical values.
b) To compute the test statistic, we need to calculate the standard error and the t-value. The formula for the standard error (SE) is as follows:
SE = √(sample variance / sample size) = √(9 / 10) = √0.9
The formula for the t-value is:
t-value = (sample mean - hypothesized mean) / SE = (8.9 - 15) / √0.9
To find the critical value, we need to find the t-value at the given α level and degrees of freedom (df). Since the sample size is 10, the degrees of freedom is 10 - 1 = 9.
Using a t-table or t-distribution calculator, we can find the critical value at α = 0.02 and df = 9.
c) The p-value represents the probability of observing a test statistic as extreme or more extreme than the one obtained, assuming the null hypothesis is true.
To calculate the p-value, we need to find the probability that the absolute value of the t-value is greater than or equal to the observed test statistic.
d) Based on the obtained test statistic, critical value, and p-value, we can make a conclusion about the hypothesis test. This conclusion will depend on whether the observed test statistic falls within the critical region or if the p-value is less than the chosen significance level (α).
a) The rejection rule for a hypothesis test at α = 0.02 can be determined by comparing the p-value to the significance level. In this case, α = 0.02 means that we want to reject the null hypothesis if the p-value is less than 0.02. So, the rejection rule is: Reject H0 if p-value < 0.02.
b) To compute the value of the test statistic, we first need to calculate the standard error of the mean (SEM). The formula for SEM is: SEM = √(population variance / sample size).
Given that the population variance is 9 and the sample size is 10, we can calculate the SEM:
SEM = √(9 / 10) = √0.9 ≈ 0.95
Next, we can calculate the test statistic (z-score) using the formula: (sample mean - hypothesized mean) / SEM
Test statistic = (8.9 - 15) / 0.95 ≈ -6.84
To find the critical value, we need to determine the z-score that corresponds to the significance level α = 0.02. The critical value is determined using a standard normal distribution table or a statistical software. For α = 0.02, the critical value is approximately ±2.575 (assuming a two-tailed test).
c) The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true. In this case, the p-value represents the probability of a sample mean of 8.9 or lower, or 8.9 or higher, given that µ is actually equal to 15.
To find the p-value, we need to calculate the area in the tail(s) of the distribution that is more extreme than the observed test statistic. Since this is a two-tailed test, we need to find the area in both tails.
Using a standard normal distribution table or a statistical software, we find that the p-value for a test statistic of -6.84 is extremely small, almost 0.
d) The conclusion can be drawn based on the results of the hypothesis test and the given significance level.
Since the p-value (approximately 0) is less than the significance level α = 0.01, we reject the null hypothesis. This means that there is strong evidence against the claim that the population mean (µ) is equal to 15. Instead, we have enough evidence to support the alternative hypothesis, which states that the population mean is not equal to 15.