Question

what i have learned
activity 2
a. find the values for each of the following degree of freedom and t - values with the given percentile. write your answer on the space provided.

1. $t_{\alpha/2}$ and $n = 16$ for the 99% confidence interval

$df = $ ____ $t_{(0.005,15)} = $ ____

1. $t_{\alpha/2}$ and $n = 25$ for the 98% confidence interval

$df = $ ____ $t_{(0.01,24)} = $ ____

1. $t_{\alpha/2}$ and $n = 8$ for the 95% confidence interval

$df = $ ____ $t_{(0.025,7)} = $ ____

1. $t_{\alpha/2}$ and $n = 12$ for the 90% confidence interval

$df = $ ____ $t_{(0.05,11)} = $ ____

1. $t_{\alpha/2}$ and $n = 20$ for the 99% confidence interval

$df = $ ____ $t_{(0.005,19)} = $ ____
b. summarize the lesson by completing the sentences below:

1. when asked of an estimation of population mean but population standard deviation is unknown, the ______ can be used.
2. the t - distribution is similar to the standard normal distribution in the following ways:

2.1 ______
2.2 ______
2.3 ______
2.4 ______

Answer:

Part A: Finding t - values and Degrees of Freedom

1. \( t_{\alpha/2} \) and \( n = 16 \) for 99% confidence interval

Step 1: Calculate degrees of freedom (\( df \))

The formula for degrees of freedom for a t - distribution when dealing with a sample mean is \( df=n - 1 \). Given \( n = 16 \), we have \( df=16 - 1=15 \).

Step 2: Determine the significance level (\( \alpha \)) and \( \alpha/2 \)

For a 99% confidence interval, the confidence level \( CL = 0.99 \). The significance level \( \alpha=1 - CL=1 - 0.99 = 0.01 \). Then \( \alpha/2=\frac{0.01}{2}=0.005 \).

Step 3: Find the t - value \( t_{\alpha/2,df} \)

We need to find \( t_{0.005,15} \). Using a t - distribution table or a statistical software, \( t_{0.005,15}\approx2.947 \).

2. \( t_{\alpha/2} \) and \( n = 25 \) for 98% confidence interval

Step 1: Calculate degrees of freedom (\( df \))

Using \( df=n - 1 \) with \( n = 25 \), we get \( df=25 - 1 = 24 \).

Step 2: Determine the significance level (\( \alpha \)) and \( \alpha/2 \)

For a 98% confidence interval, \( CL = 0.98 \), so \( \alpha=1 - 0.98=0.02 \). Then \( \alpha/2=\frac{0.02}{2}=0.01 \).

Step 3: Find the t - value \( t_{\alpha/2,df} \)

We need to find \( t_{0.01,24} \). Using a t - distribution table or software, \( t_{0.01,24}\approx2.492 \).

3. \( t_{\alpha/2} \) and \( n = 8 \) for 95% confidence interval

Step 1: Calculate degrees of freedom (\( df \))

Using \( df=n - 1 \) with \( n = 8 \), we get \( df=8 - 1 = 7 \).

Step 2: Determine the significance level (\( \alpha \)) and \( \alpha/2 \)

For a 95% confidence interval, \( CL = 0.95 \), so \( \alpha=1 - 0.95 = 0.05 \). Then \( \alpha/2=\frac{0.05}{2}=0.025 \).

Step 3: Find the t - value \( t_{\alpha/2,df} \)

We need to find \( t_{0.025,7} \). Using a t - distribution table or software, \( t_{0.025,7}\approx2.365 \).

4. \( t_{\alpha/2} \) and \( n = 12 \) for 90% confidence interval

Step 1: Calculate degrees of freedom (\( df \))

Using \( df=n - 1 \) with \( n = 12 \), we get \( df=12 - 1 = 11 \).

Step 2: Determine the significance level (\( \alpha \)) and \( \alpha/2 \)

For a 90% confidence interval, \( CL = 0.90 \), so \( \alpha=1 - 0.90 = 0.10 \). Then \( \alpha/2=\frac{0.10}{2}=0.05 \).

Step 3: Find the t - value \( t_{\alpha/2,df} \)

We need to find \( t_{0.05,11} \). Using a t - distribution table or software, \( t_{0.05,11}\approx1.796 \).

5. \( t_{\alpha/2} \) and \( n = 20 \) for 99% confidence interval

Step 1: Calculate degrees of freedom (\( df \))

Using \( df=n - 1 \) with \( n = 20 \), we get \( df=20 - 1 = 19 \).

Step 2: Determine the significance level (\( \alpha \)) and \( \alpha/2 \)

For a 99% confidence interval, \( CL = 0.99 \), so \( \alpha=1 - 0.99 = 0.01 \). Then \( \alpha/2=\frac{0.01}{2}=0.005 \).

Step 3: Find the t - value \( t_{\alpha/2,df} \)

We need to find \( t_{0.005,19} \). Using a t - distribution table or software, \( t_{0.005,19}\approx2.861 \).

Part B: Summarizing the Lesson

1. When asked for an estimation of the population mean but the population standard deviation is unknown, the t - distribution can be used.

1. The t - distribution is similar to the standard normal distribution in the following ways:

2.1 Both are symmetric about the mean (the mean of both the t - distribution and the standard normal distribution is 0).
2.2 As the degrees of freedom of the t - distribution increase, the t - distribution approaches the standard normal distribution.
2.3 Both are bell - shaped curves.
2.4 The total area under both the t - distribution curve and the standard normal distribution curve is 1.

Final Answers for Part A

1. \( df=\boldsymbol{15} \), \( t_{(0.005,15)}=\boldsymbol{2.947} \)
2. \( df=\boldsymbol{24} \), \( t_{(0.01,24)}=\boldsymbol{2.492} \)
3. \( df=\boldsymbol{7} \), \( t_{(0.025,7)}=\boldsymbol{2.365} \)
4. \( df=\boldsymbol{11} \), \( t_{(0.05,11)}=\boldsymbol{1.796} \)
5. \( df=\boldsymbol{19} \), \( t_{(0.005,19)}=\boldsymbol{2.861} \)

For Part B, the answers are as summarized above.