1. At the national level, the average income for a given occupation is $74 914. You want to determine if the income for member of the occupation is the same in Chicago. You take a random sample of 112 members of the occupation in Chicago. You find that mean of your sample is $78 695. Assume that the population standard deviation is $14 530.
(a) State the hypotheses and test them with the critical value approach and a
significance level of α = 0.05.
(b)What is the p-value of the test statistic?
(c)What is the 95% confidence interval for μ?
2. Consider the following data:
23 17 20 29 21 14 19 24
Develop and test a hypothesis to determine if the variance is greater than 4
I am very confused on these two questions, please help somehow
Here is a beginning:
1. Z = (score-mean)/SD = (78695-74914)/14530
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of that Z score.
2. Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Sure, I can help you with these questions. Let's start with question 1.
(a) To determine if the income for members of the occupation in Chicago is the same as the national average, we can set up the following hypotheses:
Null hypothesis (H0): The mean income for members of the occupation in Chicago is equal to the national average.
Alternative hypothesis (Ha): The mean income for members of the occupation in Chicago is different from the national average.
To test these hypotheses using the critical value approach and a significance level of α = 0.05, we can use a two-tailed t-test since the sample size is relatively small (< 30) and we do not know the population standard deviation.
The test statistic formula for a two-tailed t-test is: t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Substituting the given values:
Sample mean (x̄) = $78,695
Population mean (μ) = $74,914
Population standard deviation (σ) = $14,530
Sample size (n) = 112
Calculating the t-value: t = ($78,695 - $74,914) / ($14,530 / sqrt(112))
To compare this t-value to the critical values, we need to determine the degrees of freedom. Since we are using the sample standard deviation, we calculate the degrees of freedom as (n - 1).
Degrees of freedom (df) = 112 - 1 = 111
Using a t-distribution table or a statistical software, find the critical values at a significance level of α/2 = 0.05/2 = 0.025 for a two-tailed test with 111 degrees of freedom.
(b) The p-value of the test statistic can be calculated using the t-distribution. The p-value tells us the probability of observing a sample mean as extreme as the one we obtained, assuming the null hypothesis is true. We can compare this p-value to our significance level of α = 0.05 to determine if we reject or fail to reject the null hypothesis.
To calculate the p-value, we need to find the probability of getting a t-value as extreme or more extreme than the one we obtained in our sample.
(c) The 95% confidence interval for μ can be calculated using the formula:
Confidence interval = sample mean ± (critical value * standard error)
The critical value can be obtained from the t-distribution table or a statistical software at a significance level of α/2 = 0.05/2 = 0.025, and the standard error is calculated as the sample standard deviation divided by the square root of the sample size.
Now, let's move on to question 2.
To test if the variance is greater than 4 for the given data set:
Data: 23, 17, 20, 29, 21, 14, 19, 24
We can set up the following hypotheses:
Null hypothesis (H0): The variance of the data set is equal to 4.
Alternative hypothesis (Ha): The variance of the data set is greater than 4.
To test these hypotheses, we can use the chi-square test for variance. The test statistic follows a chi-square distribution with (n-1) degrees of freedom, where n is the sample size.
The test statistic formula for the chi-square test for variance is: χ² = (n-1) * (sample standard deviation ^2) / (desired variance)
Substituting the given values:
Sample size (n) = 8
Sample standard deviation (s) = calculated from the given data
Desired variance (σ²) = 4
Calculating the chi-square value: χ² = (8-1) * (sample standard deviation^2) / 4
To compare this chi-square value to the critical value, we need to determine the degrees of freedom which is (n-1) = 8-1 = 7. Using a chi-square distribution table or a statistical software, find the critical value based on the desired significance level.
I hope this explanation helps you understand how to approach these two questions. If you have any more questions or need further clarification, please let me know!