Hello I need help solving a statistics problem and have no idea where to start. Can anyone help?
In a recent national survey, the mean weekly allowance for a nine-year-old child from his or her parents was reported to be $3.65. A random sample of 45 nine-year-olds in northwestern Ohio revealed the mean allowance to be $3.69 with a standard deviation of .24. At the .05 level of significance, is there a difference in the mean allowances nationally and the mean allowances in northwestern Ohio for nine-year-olds?
Try a one-sample z-test for your problem.
Using the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (3.69 - 3.65)/(.24/√45)
Finish the calculation. Use a z-table to determine the critical or cutoff value to reject or fail to reject the null hypothesis at .05 level of significance. This will be a two-tailed test because the problem is just asking if there is a difference. Keep that in mind when you are looking at the value in the table.
Once you have the value from the table, compare to the test statistic. If the test statistic exceeds the value from the table, reject the null and conclude a difference. If the test statistic does not exceed the value from the table, fail to reject the null and you cannot conclude a difference.
I hope this will help get you started on your problem.
Certainly, I can help you solve the statistics problem step by step.
To determine whether there is a significant difference between the mean allowances nationally and in northwestern Ohio for nine-year-olds, you can conduct a hypothesis test. Here's how you can approach it:
Step 1: Identify the null and alternative hypotheses.
- Null hypothesis (H0): There is no difference between the mean allowances nationally and in northwestern Ohio for nine-year-olds.
- Alternative hypothesis (HA): There is a difference between the mean allowances nationally and in northwestern Ohio for nine-year-olds.
Step 2: Choose the appropriate statistical test.
Since you have the mean and standard deviation of a sample, and you want to compare it with a known population mean, you can use a one-sample t-test.
Step 3: Set the level of significance.
In this case, the level of significance is given as 0.05.
Step 4: Calculate the test statistic.
The formula for the one-sample t-test is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ : sample mean
μ : population mean under the null hypothesis
s : sample standard deviation
n : sample size
For this problem:
x̄ = $3.69
μ = $3.65
s = 0.24
n = 45
Substituting the values:
t = (3.69 - 3.65) / (0.24 / √45)
Step 5: Determine the critical value.
To determine the critical value, you need to refer to the t-distribution table or use statistical software. Since the sample size is large (n > 30), the t-distribution can be approximated to the standard normal distribution. For a significance level of 0.05, the critical value for a two-tailed test is approximately ±1.96.
Step 6: Compare the test statistic with the critical value.
If the test statistic falls within the critical region (outside the range of -1.96 to 1.96), you reject the null hypothesis. Otherwise, you fail to reject the null hypothesis.
Step 7: Draw a conclusion.
Based on the test statistic comparing the national mean and mean allowance in northwestern Ohio, and comparing it with the critical value, you can determine whether to reject or fail to reject the null hypothesis. If you reject the null hypothesis, it means there is a significant difference between the means of the two groups. If you fail to reject the null hypothesis, it means there is no significant difference.
I hope this helps you get started on solving the problem. Let me know if you have any further questions!