the average length of time for students to register for fall classes at a certain college has been 50 minutes. a new registration procedure using modern computing machine is being tried. if a random sample of 12 students had an average registration time of 42 minutes with a standard deviation of 11.90 minutes under the new system, test the hypothesis that the population mean is now less than 50 using:a. 0.05 level of significance and b. 0.01 level of significance.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
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.
To test the hypothesis that the population mean is now less than 50, we can use a one-sample t-test.
Let's define our null and alternative hypotheses:
Null hypothesis (H0): The population mean (μ) is equal to 50.
Alternative hypothesis (HA): The population mean (μ) is less than 50.
a. Testing at a 0.05 level of significance:
Step 1: Calculate the test statistic
The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean,
μ is the population mean (assumed to be 50),
s is the sample standard deviation,
n is the sample size.
Using the values given:
x̄ = 42 minutes,
μ = 50 minutes,
s = 11.90 minutes,
n = 12 students.
t = (42 - 50) / (11.90 / √12) = -2.03
Step 2: Determine the critical value
Since we want to test if the population mean is less than 50, we will use a one-tailed test. With a sample size of 12, degrees of freedom (df) is 11. From the t-distribution table, the critical value at a 0.05 level of significance is approximately -1.795.
Step 3: Compare the test statistic with the critical value
Since the calculated t-value (-2.03) is less than the critical value (-1.795), we reject the null hypothesis.
b. Testing at a 0.01 level of significance:
Repeat the steps above with a significance level of 0.01:
Step 1: Calculate the test statistic
Using the same values as before: t = -2.03
Step 2: Determine the critical value
From the t-distribution table, the critical value at a 0.01 level of significance is approximately -2.718.
Step 3: Compare the test statistic with the critical value
The calculated t-value (-2.03) is greater than the critical value (-2.718), so we fail to reject the null hypothesis.
In conclusion, at a 0.05 level of significance, we reject the hypothesis that the population mean is now less than 50 minutes. However, at a 0.01 level of significance, we fail to reject the null hypothesis.
To test the hypothesis that the population mean registration time is less than 50 minutes using the new registration procedure, we can conduct a one-sample t-test. We will compare the sample mean to the hypothesized population mean of 50 minutes. Here's how you can do it:
a. Testing at a 0.05 level of significance:
1. State the hypotheses:
- Null hypothesis (H0): The population mean registration time is equal to 50 minutes.
- Alternative hypothesis (Ha): The population mean registration time is less than 50 minutes.
2. Calculate the test statistic:
- The test statistic t can be calculated using the formula: t = (sample mean - hypothesized mean) / (sample standard deviation/sqrt(sample size))
- In this case, the sample mean is 42 minutes, the hypothesized mean is 50 minutes, the sample standard deviation is 11.90 minutes, and the sample size is 12.
- Therefore, t = (42 - 50) / (11.90 / sqrt(12)) = -2.692
3. Determine the critical value:
- Since we are testing at a significance level of 0.05, we need to find the critical value for a one-tailed test at a 0.05 level of significance with degrees of freedom (df) = sample size - 1.
- From a t-distribution table or using statistical software, the critical value for a t-test with df = 11 at a 0.05 level of significance is approximately -1.796.
4. Make a decision:
- If the test statistic is less than the critical value (-2.692 < -1.796), we reject the null hypothesis.
- Therefore, we reject the hypothesis that the population mean registration time is equal to 50 minutes at a 0.05 level of significance. There is evidence to suggest that the population mean registration time is less than 50 minutes under the new system.
b. Testing at a 0.01 level of significance:
1. State the hypotheses:
- Null hypothesis (H0): The population mean registration time is equal to 50 minutes.
- Alternative hypothesis (Ha): The population mean registration time is less than 50 minutes.
2. Calculate the test statistic:
- Same as in part a, t = -2.692.
3. Determine the critical value:
- At a significance level of 0.01 and df = 11, the critical value for a one-tailed test is approximately -2.718.
4. Make a decision:
- If the test statistic is less than the critical value (-2.692 < -2.718), we reject the null hypothesis.
- Thus, we reject the hypothesis that the population mean registration time is equal to 50 minutes at a 0.01 level of significance. There is evidence to suggest that the population mean registration time is less than 50 minutes under the new system.
Note: Always double-check critical values from a t-distribution table or use statistical software to ensure accuracy.