Question

variables are normally distributed and the variances are unequal. whiting franklin mean age (in years) 61.6 59.7 standard deviation (in years) 5.4 3.9 use \\( \mu _ { 1 } \\) for the average age of houses in whiting. part: 0 / 5 part 1 of 5 (a) state the hypotheses and identify the claim. \\( h _ { 0 } : \\) \\( h _ { 1 } : \\) this hypothesis test is a test.

Explanation:

Step1: Determine the null hypothesis

The null hypothesis \( H_0 \) typically assumes no difference or equality. So \( H_0: \mu_1 = \mu_2 \) (not claim).

Step2: Determine the alternative hypothesis

Since we are comparing the average age of houses in Whiting (\( \mu_1 \)) and Franklin (\( \mu_2 \)), and likely testing if \( \mu_1 \) is different (or in a direction) from \( \mu_2 \). Assuming a two - tailed or a one - tailed test, but from the context of comparing two means, if we assume a two - tailed test for difference, \( H_1: \mu_1
eq \mu_2 \) (claim). But if we assume a one - tailed test, say \( \mu_1>\mu_2 \), \( H_1: \mu_1 > \mu_2 \) (claim). But from the given data, the mean of Whiting (61.6) is greater than Franklin (59.7). Let's assume we are testing if \( \mu_1
eq\mu_2 \) (two - tailed) or \( \mu_1 > \mu_2 \) (one - tailed). Let's go with the most common case of testing for a difference. So \( H_0: \mu_1=\mu_2 \) (not claim), \( H_1: \mu_1
eq\mu_2 \) (claim) or \( H_1: \mu_1 > \mu_2 \) (claim).

For the hypothesis test type, since we are comparing two means with unequal variances (given in the problem statement: "variables are normally distributed and the variances are unequal"), this is a two - sample t - test (Welch's t - test) for two independent samples.

Answer:

\( H_0: \boldsymbol{\mu_1=\mu_2} \) (not claim), \( H_1: \boldsymbol{\mu_1
eq\mu_2} \) (claim) (or \( H_1: \boldsymbol{\mu_1 > \mu_2} \) if one - tailed). This hypothesis test is a \(\boldsymbol{two - sample\ t - test}\) (Welch's t - test) for two independent samples.