AP Statistics Quiz 10.2

1.

Mrs. Toguchi wants to determine if politicians tend to have higher salaries, on average, than first

responders (specifically emergency medical technicians, EMTs). Independent random samples from each

of these two professions were selected and their salaries were recorded. The data are summarized below.

a) the distribution of salary in each group is skewed right. The use of two-sample t procedures is still justified why?

b) construct and interpret a 99% confidence interval for the difference in the true mean salaries for politicians and EMTs

Answer:

Hello some data related to your question is missing attached below is the missing data

a) we will reject the null hypothesis hence it can be determined that the salary of politicians is greater than that of EMTs through the use of two-sample t procedures. since ( to > critical value )

b) ( 3.819 , 8.5226 )

Step-by-step explanation:

Applying the Given data:

The test hypothesis :

H0 : μ1 = μ2

Ha : μ1 > 2

A) since ( to > critical value ) we will reject the null hypothesis hence it can be determined that the salary of politicians is greater than that of EMTs through the use of two-sample t procedures

B ) construct and interpret

next calculate the SP value

[tex]S^{2} _{p}[/tex] = [tex]\frac{( n1 - 1 )S^{2} _{1} + ( n_{2}- 1 )S^{2} _{2} }{n1 + n2 - 2}[/tex]

where: S1 = 7524 , n1 = 50 , n2 = 100 , S2 = 5551   ( input values into equation above )

∴ [tex]S^{2} _{p}[/tex] = 39354491.3716

hence : [tex]S_{p}[/tex] = 6273.32

next calculate the value of test statistic

to = ( X1 - X2 - Δo ) / Sp [tex]\sqrt{\frac{1}{n1} + \frac{1}{n2} }[/tex]

where: X1 = 61234 , X2 = 54529 , Sp = 6273.32 , n1 = 50 , n2 = 100  ( input values into equation above )

to = 6705 / 1086.5702 = 6.1708

Df ( degree of freedom ) = n1 + n2 - 2 = 148

critical value at ∝ = 0.01  = ± 2.3518  ( using excel function : T.INV(0.01,148) )

hence the confidence Interval

= ( 6.1708 ± 2.3518 )

= ( 3.819 , 8.5226 )

For the table of Mrs. Toguchi the t test for two sample is not justified and the confidence interval for this two values is (6.1708, ±2.3518).

Random sample is the way to choose a number or sample in such a manner that each of the sample of the group has an equal probability to be chosen.

Mrs. Toguchi wants to determine if politicians tend to have higher salaries, on average, than first responders (specifically emergency medical technicians, EMTs).

Independent random samples from each of these two professions were selected and their salaries were recorded. The data are summarized below.

                                   Politicians        EMT's

Mean                            $61,234       $54,529

Standard deviation      $7524          $5551

n                                      50                100

The hypothesis for the test,

[tex]H_0:\mu_1=\mu_2\\H_a:\mu_1 > 2[/tex]

No it is not justifies, as the null hypothesis is rejected.

The pooled estimate of a common population standard deviation can be given as,

[tex]S_p=\sqrt{\dfrac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}}[/tex]

Put the values in the above formula from the table, we get,

[tex]S_p=\sqrt{\dfrac{(50-1)(7524)^2+(100-1)(5551)^2}{50+100-2}}\\S_p=6273.32[/tex]

Use the formula of test statistics for equal variance as,

[tex]t=\dfrac{\overline y_1-\overline y_2-0}{s_p\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}[/tex]

Put the values from the table as,

[tex]t=\dfrac{61234-54529-0}{(6263.32)\sqrt{\dfrac{1}{50}+\dfrac{1}{100}}}\\t=6.1708[/tex]

The degree of freedom can be given as,

[tex]DF=100+50-2\\DF=148[/tex]

The critical value for 99 percent interval (0.01) is ±2.3518.

Hence, the confidence interval for this two values is (6.1708, ±2.3518).

Thus, for the table of Mrs. Toguchi the t test for two sample is not justified and the confidence interval for this two values is (6.1708, ±2.3518).

Learn more about the random sample here;

https://brainly.com/question/17831271

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