Question

listed in the accompanying table are weights (kg) of randomly selected u s army male personnel measured in 1988 (from \ansur | 1988\) and different weights (kg) of randomly selected u s army male personnel measured in 2012 (from \ansur ii 2012\). assume that the two samples are independent simple random samples selected from normally distributed populations do not assume that the population standard deviations are equal. complete parts (a) and (b)
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a. use a 0.01 significance level to test the claim that the mean weight of the 1988 popula
what are the null and alternative hypotheses? assume that population 1 consists of the
a. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}<mu_{2} )
c. ( h_{0}: mu_{1} leq mu_{2} )
( h_{1}: mu_{1}>mu_{2} )
the test statistic is -0.57 (round to two decimal places as needed.)
the p - value is (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The formula for degrees of freedom for two - sample \(t\) - test (equal variances) is \(df=n_1 + n_2-2\).
Let \(n_1\) be the sample size of 2012 data and \(n_2\) be the sample size of 1988 data. Counting the data points, \(n_1 = 15\), \(n_2=12\). So \(df=15 + 12-2=25\).

Step2: Find the P - value

Since the test statistic \(t=-0.57\) and it is a left - tailed test (\(H_1:\mu_1<\mu_2\)).
Using a \(t\) - distribution table or a calculator with the \(t\) - distribution function \(P(T < t)\) where \(T\) follows a \(t\) - distribution with \(df = 25\) and \(t=-0.57\).
Using a calculator (e.g., in R: \(pt(-0.57,25)\)), we get \(P\approx0.287\)

Answer:

\(0.287\)