For a random sample of 12 observations, the sample mean was 407 and the standard deviation was 6. Using the .05 significant level, answer the following questions.
What would the t value be?
What is the decision rule?
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I think the t-value would be t = -3.0845
Any help would be great.
To determine the t-value, we need to calculate the standard error (SE). The formula for SE is:
SE = standard deviation / square root of sample size
Given that the standard deviation is 6 and the sample size is 12, we can calculate the SE as follows:
SE = 6 / √12 ≈ 1.732
Now, let's find the t-value by dividing the difference between the sample mean and the hypothesized population mean by the standard error:
t = (sample mean - hypothesized population mean) / SE
Here, the hypothesized population mean is not given, so we cannot calculate the exact t-value. However, if you have a hypothesized population mean, you can substitute it into the formula to find the t-value.
Regarding the decision rule, we need to consider the significance level, which is given as 0.05 in this case. Typically, for a two-tailed test at a significance level of 0.05, the critical t-values are determined by dividing the significance level by 2 and looking up the t-value in a t-table with the appropriate degrees of freedom.
Since the sample size is 12, the degrees of freedom for a t-distribution is calculated as (sample size - 1) = (12 - 1) = 11. Looking up the critical t-value for a two-tailed test with 11 degrees of freedom and a significance level of 0.025 (0.05/2), we would find the critical t-value in the t-table.
I'm sorry that I couldn't provide the exact t-value and decision rule without a specific hypothesized population mean, but I hope this explanation helps you understand how to calculate them. If you have a hypothesized population mean or any other specific information, please provide it and I'll be happy to assist you further.