You are conducting an experiment to see if a given therapy works to reduce test
anxiety in a sample of college students. A standard measure of test anxiety is known to
produce a � = 19. In the sample you draw of 86 the mean is equal 17.5 with standard
deviation s = 10.
a- Use an � level of 0.05 to test the hypothesis that the mean is less than 19.
b- Find the 95 % one sample con�dence interval for the mean.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. Remember that your hypothesis leads to a one-tailed test.
95% = mean ± 1.96 SEm
a- To test the hypothesis that the mean is less than 19, we can use a one-sample t-test. The null hypothesis (H0) is that the mean is equal to 19, and the alternative hypothesis (Ha) is that the mean is less than 19.
Our sample size is 86, with a sample mean (x̄) of 17.5 and a sample standard deviation (s) of 10.
To calculate the t-statistic, we use the formula:
t = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ = sample mean
μ = population mean (hypothesized mean)
s = sample standard deviation
n = sample size
Plugging in the values:
t = (17.5 - 19) / (10 / sqrt(86))
t = -1.5 / (10 / sqrt(86))
We can now find the critical t-value for our test. Since we want to test if the mean is less than 19, and our level of significance (α) is 0.05, we find the critical t-value using the t-distribution table or a statistical software. From the table, the critical t-value for a one-tailed test with degrees of freedom (df) = 85 and a significance level of 0.05 is approximately -1.663.
Now, let's compare the calculated t-value to the critical t-value:
-1.5 / (10 / sqrt(86)) < -1.663
Since the calculated t-value is not less than the critical t-value, we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean test anxiety score is less than 19 at the 0.05 level of significance.
b- To find the 95% confidence interval for the mean, we can use the formula:
Confidence interval = (x̄ - t * (s / √n), x̄ + t * (s / √n))
Where:
x̄ = sample mean
t = critical t-value (corresponding to the confidence level and degrees of freedom)
s = sample standard deviation
n = sample size
Since we want a 95% confidence interval, the corresponding critical t-value can be found using the t-distribution table or statistical software. For a two-tailed test with df = 85, the critical t-value is approximately 1.986.
Plugging in the values:
Confidence interval = (17.5 - 1.986 * (10 / sqrt(86)), 17.5 + 1.986 * (10 / sqrt(86)))
Confidence interval = (17.5 - 1.986 * 1.078, 17.5 + 1.986 * 1.078)
Confidence interval = (15.39, 19.61)
The 95% confidence interval for the mean test anxiety score is (15.39, 19.61). This means that we can be 95% confident that the true population mean falls within this range.
To test the hypothesis and calculate the confidence interval, you can follow these steps:
a) Testing the Hypothesis:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H₀): The mean test anxiety in the population is equal to 19.
- Alternative hypothesis (H₁): The mean test anxiety in the population is less than 19.
Step 2: Choose the significance level (α):
The given significance level is α = 0.05.
Step 3: Calculate the test statistic:
The test statistic to use in this case is the t-statistic since the population standard deviation (σ) is unknown.
The formula to calculate the t-statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)
Given information:
Sample mean (x̄) = 17.5
Hypothesized mean (μ₀) = 19
Sample standard deviation (s) = 10
Sample size (n) = 86
Substituting the values into the formula:
t = (17.5 - 19) / (10 / √86)
Step 4: Determine the critical region:
Since the alternative hypothesis is one-sided (less than 19), we need to find the critical t-value for a one-tailed test with α = 0.05 and degrees of freedom (df) equal to n - 1, where n is the sample size.
Using a t-table or a statistical calculator, the critical t-value for α = 0.05 and df = 85 is approximately -1.663.
Step 5: Make a decision:
If the calculated t-value falls within the critical region, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
b) Confidence Interval Calculation:
Step 1: Determine the critical value for the desired confidence level (95% confidence level):
Since it's a two-tailed test for the confidence interval, we need to find the critical t-value for α/2 = 0.025, with df = n - 1.
Using a t-table or a statistical calculator, the critical t-value for α/2 = 0.025 and df = 85 is approximately ±1.992.
Step 2: Calculate the margin of error:
The margin of error is determined by multiplying the critical value by the standard error, where the standard error (SE) is calculated as:
SE = (sample standard deviation) / √sample size
Given information:
Sample standard deviation (s) = 10
Sample size (n) = 86
Substituting the values into the formula:
SE = 10 / √86
Step 3: Calculate the confidence interval:
The confidence interval is determined by subtracting and adding the margin of error to the sample mean.
Lower confidence limit = sample mean - margin of error
Upper confidence limit = sample mean + margin of error
Substituting the values into the formula:
Lower confidence limit = 17.5 - (1.992 * (10 / √86))
Upper confidence limit = 17.5 + (1.992 * (10 / √86))
These calculations will provide you with both the hypothesis test result and the confidence interval for the given scenario.