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.

1. the t - distribution is similar to the standard normal distribution in the

following ways:
2.1 __________
2.2 __________
2.3 __________
2.4 __________

Answer:

the t - distribution table or a statistical calculator, for \( df = 15 \) and \( \alpha/2=0.005 \), the t - value \( t_{0.005,15}\approx2.947 \) (this value can be obtained from t - distribution tables where the row is \( df = 15 \) and the column is for two - tailed test with \( \alpha = 0.01 \) (or one - tailed \( \alpha/2=0.005 \))).

So, \( df = 15 \), \( t_{(0.005,15)}\approx2.947 \)

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

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

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

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

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

Step 3: Find the t - value \( t_{(0.01,24)} \)

From the t - distribution table, for \( df = 24 \) and \( \alpha/2 = 0.01 \), the t - value \( t_{0.01,24}\approx2.492 \) (looking at the row for \( df = 24 \) and the column for two - tailed test with \( \alpha=0.02 \) (or one - tailed \( \alpha/2 = 0.01 \))).

So, \( df = 24 \), \( t_{(0.01,24)}\approx2.492 \)

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 have \( df=8 - 1 = 7 \).

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

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

Step 3: Find the t - value \( t_{(0.025,7)} \)

From the t - distribution table, for \( df = 7 \) and \( \alpha/2=0.025 \), the t - value \( t_{0.025,7}\approx2.365 \) (row for \( df = 7 \), column for two - tailed test with \( \alpha = 0.05 \)).

So, \( df = 7 \), \( t_{(0.025,7)}\approx2.365 \)

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, \( \alpha=1 - 0.90 = 0.10 \), and \( \alpha/2=\frac{0.10}{2}=0.05 \).

Step 3: Find the t - value \( t_{(0.05,11)} \)

From the t - distribution table, for \( df = 11 \) and \( \alpha/2 = 0.05 \), the t - value \( t_{0.05,11}\approx1.796 \) (row for \( df = 11 \), column for two - tailed test with \( \alpha=0.10 \)).

So, \( df = 11 \), \( t_{(0.05,11)}\approx1.796 \)

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 have \( df=20 - 1 = 19 \).

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

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

Step 3: Find the t - value \( t_{(0.005,19)} \)

From the t - distribution table, for \( df = 19 \) and \( \alpha/2=0.005 \), the t - value \( t_{0.005,19}\approx2.861 \) (row for \( df = 19 \), column for two - tailed test with \( \alpha = 0.01 \)).

So, \( df = 19 \), \( t_{(0.005,19)}\approx2.861 \)

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 the t - distribution is 0, same as the standard normal distribution).
• 2.2 As the degrees of freedom increase, the t - distribution approaches the standard normal distribution.
• 2.3 Both are bell - shaped curves.
• 2.4 The total area under both curves is equal to 1 (representing a probability distribution).

Final Answers for Part A:

1. \( df=\boldsymbol{15} \), \( t_{(0.005,15)}\approx\boldsymbol{2.947} \)

1. \( df=\boldsymbol{24} \), \( t_{(0.01,24)}\approx\boldsymbol{2.492} \)

1. \( df=\boldsymbol{7} \), \( t_{(0.025,7)}\approx\boldsymbol{2.365} \)

1. \( df=\boldsymbol{11} \), \( t_{(0.05,11)}\approx\boldsymbol{1.796} \)

1. \( df=\boldsymbol{19} \), \( t_{(0.005,19)}\approx\boldsymbol{2.861} \)

For Part B, the answers are as provided in the summarization above.