Question

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is $150,000, with a standard deviation of $40,000. A corresponding sample of 25 people at the second development had a mean of $180,000, with a standard deviation of $30,000. Assume the population standard deviations are the same.
1. State the decision rule for 0.05 significance level: H0: μ1 = μ2; H1:μ1 ≠ μ2.2. Compute the value of the test statistic

Answer 1

Answer:

The decision rule is  

  Reject the null hypothesis

The test statistics is $t = -2.699$

Step-by-step explanation:

From the question we are told that

   The first sample size is  $n_1 = 15$

   The mean at first deployment is  $\= x _1 = \ 150 000$

   The standard deviation is  $s_1 = \ 40000$

    The second sample size is $n_2 = 25$

    The mean at second deployment is  $\= x_2 = \ 180000$

    The standard deviation is  $s_2 = \ 30 000$

The null hypothesis is  $H_o : \mu_1 = \mu_2$

The alternative hypothesis is  $H_a : \mu_1 \ne \mu_2$

   The level of significance is  $\alpha = 0.05$

Generally the degree of freedom is mathematically represented as

      $df =n_1 + n_2 -2$

=>   $df =40 -2$

=>   $df =38$

Generally the pooled variance is mathematically represented as

      $s_p^2 = \frac{x(n_1 -1) s_1^2 + (n_2 - 1)s_2^2}{df}$

=>   $s_p^2 = \frac{(15 -1) 40000^2 + (25 - 1)30000^2}{38}$

=>   $s_p^2 = 1.1579 * 10^{9}$

Generally the test statistics is mathematically represented as

      $t = \frac{ \= x_1 - \= x_2 }{\sqrt{s_p [\frac{1}{n_1} + \frac{1}{n_2} ]} }$

=>   $t = \frac{ 150000 - 180000 }{\sqrt{1.11579 *10^{9} [\frac{1}{15} + \frac{1}{25} ]} }$

=>   $t = -2.699$

Generally from the  t distribution table the probability corresponding to the t statistics value to the left is  

     $t_{-2.699 , 38} = 0.00516046$

Generally the p -value is mathematically represented as

     $p-value = 2* t_{-2.699, 38}$

=>  $p-value = 2* 0.00516046$

=>  $p-value = 0.01032$

From the obtained value we see that the  $p-value < \alpha$  hence

   The decision rule is  

  Reject the null hypothesis