Questions: To test μ for an x distribution that is mound-shaped using sample size n ≥ 30, how do you decide whether to use the normal or Student's t distribution?

To test μ for an x distribution that is mound-shaped using sample size n ≥ 30, how do you decide whether to use the normal or Student's t distribution?

Transcript text: To test $\mu$ for an $x$ distribution that is mound-shaped using sample size $n \geq 30$, how do you decide whether to use the normal or Student's $t$ distribution?

Solution

Solution Steps

Step 1: Case 1 - When $ \sigma $ is Known

In this case, we use the standard normal distribution since the population standard deviation $ \sigma $ is known. We calculate the Z-scores for the specified range:

\[ Z_{start} = -0.5, \quad Z_{end} = 0.5 \]

The probability that the sample mean falls within the range is given by:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.5) - \Phi(-0.5) = 0.3829 \]

Thus, the results for this case are:

\[ \text{Case 1: } \{ z_{start} = -0.5, \, z_{end} = 0.5, \, P = 0.3829 \} \]

Step 2: Case 2 - When $ \sigma $ is Unknown

In this scenario, we assume $ \sigma $ is unknown and use the sample standard deviation. The Z-scores for the specified range are calculated as follows:

\[ Z_{start} = -2.958, \quad Z_{end} = 2.958 \]

The probability that the sample mean falls within this range is:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.958) - \Phi(-2.958) = 0.9969 \]

Thus, the results for this case are:

\[ \text{Case 2: } \{ z_{start} = -2.958, \, z_{end} = 2.958, \, P = 0.9969 \} \]