Question

determine the hypotheses. choose the correct answer below.
a. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}
eq mu_{2} )
c. ( h_{0}: mu_{1}>mu_{2} )
( h_{1}: mu_{1} leq mu_{2} )
find the test statistic.
( t_{\text {stat }}=1.56 ) (round to two decimal places as needed.)
find the p - value.
p - value ( =square ) (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

Assume equal variances and use the formula \(df = n_1 + n_2-2\). But since sample sizes \(n_1\) and \(n_2\) are not given, assume a two - tailed test (from \(H_1:\mu_1
eq\mu_2\) in option A). For a \(t\) - distribution, if we assume a large enough sample (or use a standard \(t\) - table approximation for a two - tailed test), we can use the formula for the \(p\) - value. The \(p\) - value for a two - tailed \(t\) - test with \(t_{STAT}=1.56\) is \(2\times P(T > |t_{STAT}|)\) where \(T\) follows a \(t\) - distribution.
Using a \(t\) - table or statistical software (e.g., in Excel, if we assume a large \(n\) (approximating to a standard normal for simplicity, but more accurately for a \(t\) - distribution), for a two - tailed test:
\(p - value=2\times(1 - P(T\leq1.56))\)
If we use a \(t\) - distribution with a large number of degrees of freedom (as the \(t\) - distribution approaches the standard normal as \(df
ightarrow\infty\)), \(P(T\leq1.56)\approx0.9406\) (from standard normal table or \(t\) - table for large \(df\))

Step2: Calculate the \(p\) - value

\(p - value=2\times(1 - 0.9406)=2\times0.0594 = 0.119\)

Answer:

\(0.119\)