Answer:
The confidence interval is [tex]-11.34 < \mu_1 -\mu_2 < -10.86[/tex]
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 146[/tex]
The second sample size is [tex]n_2 = 180[/tex]
The first sample mean is [tex]\= x_1 = 51.6[/tex]
The second sample mean is [tex]\= x_2 = 62.7[/tex]
The first standard deviation is [tex]\sigma _1 = 9.42[/tex]
The second standard deviation is [tex]\sigma _2 = 14.5[/tex]
Given that the confidence level is 98% then the significance level is mathematically evaluated as
[tex]\alpha = (100 -98 )\%[/tex]
[tex]\alpha = 2 \%[/tex]
[tex]\alpha = 0.02[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the z-table , the value is [tex]Z_{\frac{\alpha }{2} } = 2.33[/tex]
The reason we are obtaining critical value of
[tex]\frac{\alpha }{2}[/tex]
instead of
[tex]\alpha[/tex]
is because
[tex]\alpha[/tex]
represents the area under the normal curve where the confidence level interval (
[tex]1-\alpha[/tex]
) did not cover which include both the left and right tail while
[tex]\frac{\alpha }{2}[/tex]
is just the area of one tail which what we required to calculate the margin of error
NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\sigma_1^2}{n_1^2} + \frac{\sigma_2^2}{n_2^2} }[/tex]
substituting values
[tex]E = 2.33 * \sqrt{ \frac{9.42^2}{146^2} + \frac{14.5^2}{180^2} }[/tex]
substituting values
[tex]E = 2.33 * \sqrt{ \frac{9.42^2}{146^2} + \frac{14.5^2}{180^2} }[/tex]
[tex]E = 0.2405[/tex]
The 98% confidence interval is mathematically represented as
[tex](\= x _ 1 - \= x_2 ) - E < \mu_1 -\mu_2 < (\= x _ 1 - \= x_2 ) + E[/tex]
substituting values
[tex](51.6 - 62.7) - 0.2405 < \mu_1 -\mu_2 < (51.6 - 62.7) + 0.2405[/tex]
[tex]-11.34 < \mu_1 -\mu_2 < -10.86[/tex]
