Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population...Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.
The normal distribution has the following format: The standard deviation is: \[\sqrt{\frac{\left(\sigma_{1}\right)^{2}}{n_{1}}+\frac{\left(\sigma_{2}\right)^{2}}{n_{2}}}\nonumber...The normal distribution has the following format: The standard deviation is:√(σ1)2n1+(σ2)2n2The test statistic (z-score) is:Zc=(¯x1−¯x2)−δ0√(σ1)2n1+(σ2)2n2 At the 5% level of significance, from the sample data, there is n…
The normal distribution has the following format: The standard deviation is: \[\sqrt{\frac{\left(\sigma_{1}\right)^{2}}{n_{1}}+\frac{\left(\sigma_{2}\right)^{2}}{n_{2}}}\nonumber...The normal distribution has the following format: The standard deviation is:√(σ1)2n1+(σ2)2n2The test statistic (z-score) is:Zc=(¯x1−¯x2)−δ0√(σ1)2n1+(σ2)2n2 At the 5% level of significance, from the sample data, there is n…