Answer:
The 90% confidence interval is [tex] -0.3433< \mu_1 - \mu_2 < 2.1433[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean for men is [tex]\= x_1 = 64.5 \ years[/tex]
The sample mean for women is [tex]\= x_2 = 63.6 \ years[/tex]
The sample size for men is [tex]n_1 = 35[/tex]
The sample size for women is [tex]n_2 = 39[/tex]
The standard deviation for men is [tex]s_1 = 3.0[/tex]
The standard deviation for women is [tex]s_2 = 3.5[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645 [/tex]
Generally the standard error is mathematically represented as
[tex]SE = \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2} }[/tex]
=> [tex]SE = \sqrt{\frac{3^2 }{35} + \frac{3.5^2}{39} }[/tex]
=> [tex]SE = 0.7558[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * SE[/tex]
=> [tex]E = 1.645 * 0.7558[/tex]
=> [tex]E = 1.2433[/tex]
Generally 90% confidence interval is mathematically represented as
[tex](\= x_1 - \= x_2) -E < \mu_1 - \mu_2 < (\= x_1 - \= x_2) -E [/tex]
=> [tex](64.5 - 63.6) -1.2433< \mu_1 - \mu_2 <(64.5 - 63.6) +1.2433[/tex]
=> [tex] -0.3433< \mu_1 - \mu_2 < 2.1433[/tex]
