A survey of over 25,000 Americans aged between 18 and 24 years revealed the following: 88.1% of the 12,678 females and 84.9% of the 12,460 males had high school diplomas. a) Do the data suggest that females are more likely to graduate from high school than males? Test at a significance level of 5%.
b) Set-up a 95% confidence interval for the difference in the graduation rates between females and males.
c) State the assumptions and conditions necessary for the above inferences to hold.

(a) Yes, the data suggest that females are more likely to graduate from high school than males.

(b) A 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .

Step-by-step explanation:

We are given that a survey of over 25,000 Americans aged between 18 and 24 years revealed the following: 88.1% of the 12,678 females and 84.9% of the 12,460 males had high school diplomas.

Let = population proportion of females who had high school diplomas.

= population proportion of males who had high school diplomas.

(a) So, Null Hypothesis, :    {means that females are less or equally likely to graduate from high school than males}

Alternate Hypothesis, :     {means that females are more likely to graduate from high school than males}

The test statistics that will be used here is Two-sample z-test statistics for proportions;

                        T.S.  =    ~  N(0,1)

where, = sample proportion of females having high school diplomas = 88.1%

= sample proportion of males having high school diplomas = 84.9%

= sample of females = 12,678

= sample of males = 12,460

So, the test statistics =  

                                    =  7.428  

The value of the standardized z-test statistic is 7.428.

Now, at a 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 7.428 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that females are more likely to graduate from high school than males.

(b) Firstly, the pivotal quantity for finding the confidence interval for the difference in population proportion is given by;

                        P.Q.  =    ~  N(0,1)

where, = sample proportion of females having high school diplomas = 88.1%

= sample proportion of males having high school diplomas = 84.9%

= sample of females = 12,678

= sample of males = 12,460

Here for constructing a 95% confidence interval we have used a Two-sample z-test statistics for proportions.

So, 95% confidence interval for the difference in population proportions, () is;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                      of significance are -1.96 & 1.96}    

P(-1.96 < < 1.96) = 0.95  

P( < < ) = 0.95  

P( < () < ) = 0.95

95% confidence interval for () = [ , ]

= [ , ]

= [0.024, 0.404]

Therefore, a 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .

(c) The assumptions and conditions necessary for the above inferences to hold are;

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.