2. At a local university, a sample of 49 evening students was selected in order to determine whether
the average age of the evening students is significantly different from 21. The average age of the
students in the sample was 23 years. The population standard deviation is known to be 3.5 years.
Determine whether or not the average age of the evening students is significantly different from 21.
Use a 0.1 level of significance.
a)statement of hypothesis
Null hypothesis: μ=21
Alternative hypothesis:μ ≠21
b)Test statistic value =4
c) rejection rule - reject if IzI> zα/ 2
d) zα/ 2=1.28
e)since 4 > 1.28 , reject the null hypothesis
To determine whether the average age of the evening students is significantly different from 21, we can perform a one-sample t-test. Here are the steps to carry out this test:
Step 1: Formulate the null and alternative hypotheses:
- Null Hypothesis (H0): The average age of the evening students is equal to 21 years.
- Alternative Hypothesis (Ha): The average age of the evening students is significantly different from 21 years.
Step 2: Set the significance level (α):
The significance level is given as 0.1, which means we will reject the null hypothesis if the p-value is less than 0.1.
Step 3: Calculate the test statistic:
We can calculate the test statistic using the formula: t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
In this case, the sample mean is 23, the population mean is 21, the population standard deviation is 3.5, and the sample size is 49.
t = (23 - 21) / (3.5 / sqrt(49))
Step 4: Find the p-value:
Using the t-distribution table or statistical software, we can find the p-value associated with the test statistic. Since the alternative hypothesis is two-sided, we will look for the p-value that is twice the probability of obtaining a t-value as extreme as the calculated t-value.
Step 5: Compare the p-value with the significance level:
If the p-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Note: Since we do not have the actual test statistics or p-value values, we cannot determine the result in this specific case. However, by following the steps outlined above, you can calculate the test statistic, find the p-value, and compare it with the significance level to determine whether the average age of the evening students is significantly different from 21.
To determine whether the average age of the evening students is significantly different from 21, we can use a hypothesis test.
Step 1: Formulate the null and alternative hypotheses.
- The null hypothesis (H0) states that the average age of the evening students is equal to 21.
- The alternative hypothesis (H1) states that the average age of the evening students is significantly different from 21.
Step 2: Select the level of significance (α). In this case, α is given as 0.1. This value represents the probability of rejecting the null hypothesis when it is true.
Step 3: Choose the appropriate statistical test. Since the population standard deviation is known and the sample size is larger than 30, we will use a Z-test.
Step 4: Calculate the test statistic. The formula for the Z-test is:
Z = (x̄ - μ) / (σ / √n)
Where:
x̄ is the sample mean (23 years),
μ is the population mean (21 years),
σ is the population standard deviation (3.5 years),
n is the sample size (49).
Plugging in the values to the formula, we get:
Z = (23 - 21) / (3.5 / √49)
Z = 2 / (3.5 / 7)
Z = 2 / 0.5
Z = 4
Step 5: Determine the critical value. Since α is 0.1 and we have a two-tailed test, we need to find the critical Z-values that would split the α value equally into the two tails. Looking up the Z-table for a two-tailed test with α = 0.1, we find the critical values to be approximately -1.645 and 1.645.
Step 6: Compare the test statistic with the critical value. The test statistic (Z = 4) is greater than the critical values (-1.645 and 1.645), so we reject the null hypothesis.
Step 7: Draw a conclusion. Since we rejected the null hypothesis, we can conclude that the average age of the evening students is significantly different from 21 at a 0.1 level of significance.
In summary, based on the provided sample data, we can conclude that the average age of the evening students is significantly different from 21 at a 0.1 level of significance.
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/probability for the Z score.