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NNADVOKAT [17]
2 years ago
5

3. The grade-point averages of 20 college seniors selected at random from a graduating class are as follows:

Mathematics
1 answer:
Ainat [17]2 years ago
6 0

The standard deviation of the grade points average of the college seniors is 0.57.

The given parameters:

3.2, 1.9 ,2.7, 2.4, 2.8, 2.9, 3.8, 3.0, 2.5, 3.3, 1.8, 2.5, 3.7, 2.8, 2.0, 3.2, 2.3, 2.1, 2.5, 1.9

The sum of the given data is calculated as follows;

∑x = 3.2 + 1.9 + 2.7 +  2.4 + 2.8 + 2.9 + 3.8 + 3.0 +  2.5 + 3.3 + 1.8 + 2.5 + 3.7 + 2.8 + 2.0 + 3.2 + 2.3 + 2.1 + 2.5 + 1.9

∑x = 53.3

The mean of the distribution is calculated as follows;

\bar x = \frac{\Sigma x}{N} \\\\\bar x = \frac{53.3}{20} \\\\\bar x = 2.665

The square of the difference between the grade point and the mean;

\Sigma (x - \bar x)^2 = (3.2- 2.665)^2 + (1.9-2.665)^2 + (2.7- 2.665)^2 + (2.4-2.665)^2  \\\\+(2.8-2.665)^2 +(2.9 - 2.665)^2 + (3.8 - 2.665)^2 + (3-2.665)^2+ \\\\(2.5 -2.665)^2 + (3.3 - 2.665)^2 + (1.8-2.665)^2 + (2.5 - 2.665)^2 + \\\\(3.7-2.665)^2 + (2.8 - 2.665)^2 + (2-2.665)^2 + (3.2-2.665)^2 + \\\\(2.3 - 2.665)^2 + (2.1-2.665)^2 + (2.5-2.665)^2 + (1.9 - 2.665)^2\\\\\Sigma (x - \bar x)^2 = 6.5055

The standard deviation of the grade points is calculated as follows;

\sigma = \sqrt{\frac{\Sigma (x-\bar x)^2}{N} } \\\\\sigma = \sqrt{\frac{6.5055}{20} } \\\\\sigma = 0.57

Thus, the standard deviation of the grade points average of the college seniors is 0.57.

Learn more about standard deviation here: brainly.com/question/12402189

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Country Financial, a financial services company, uses surveys of adults age and older to determine if personal financial fitness
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Complete Question

Country financials, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more that fair. In Feb 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair.

a

State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.

b

What is the sample proportion indicating that their  financial security was more that fair in 2012?In 2010?

c

Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?

Answer:

a

The null hypothesis is  H_o :  p_1 = p_2

The  alternative hypothesis is   H_a :  p_1  \ne  p_2

b

in 2012  \r  p_1  =0.41

in 2010  \r  p_2  =0.35  

c

The  p-value  is  p-value = 0.0072

The conclusion is

There is  sufficient evidence to conclude that the proportion of those indicating that  financial security is more fair in Feb 2010 is different from the proportion of  those indicating that financial security is more fair in Feb 2012.

Step-by-step explanation:

From the question we are told that

   The  sample size in 2012 is  n_1  =  1000

    The  number that indicated that their finance was more than fail is  k  =  410  

     The  sample  size in 2010  is  n_2  =  900

   The  number that indicated that their finance was more than fail is  u  =  315

     The  level of significance is  \alpha  =  0.05[/ex]The null hypothesis is  [tex]H_o :  p_1 = p_2

The  alternative hypothesis is   H_a :  p_1  \ne  p_2

Generally the sample proportion for  2012 is mathematically represented as

     \r  p_1  =  \frac{k}{n_1}

=>   \r  p_1  =  \frac{410}{1000}

=>   \r  p_1  =0.41

Generally the sample proportion for  2010 is mathematically represented as

      \r  p_2  =  \frac{u}{n_2}

=>   \r  p_2  =  \frac{315}{900}

=>   \r  p_2  =0.35    

Generally the pooled sample proportion is mathematically represented as

      \r p =  \frac{k + u }{n_1 + n_2}

=>     \r p =  \frac{410 + 315 }{1000+ 900}

=>      \r p =  0.3816

Generally the test statistics is mathematically represented as

     z =  \frac{( \r p_1 - \r p_2 ) - 0}{\sqrt{\r p (1 - \r p ) ( \frac{1}{n_1} +\frac{1}{n_2} )}  }

 z =  \frac{( 0.41 -0.35 ) - 0}{\sqrt{0.3816 (1 - 0.3816 ) ( \frac{1}{1000} +\frac{1}{900} )}  }    

 

 z =  \frac{( 0.41 -0.35 ) - 0}{\sqrt{0.3816 (1 - 0.3816 ) ( \frac{1}{1000} +\frac{1}{900} )}  }

   z  =  2.688

Generally the p- value  is mathematically represented as  

      p-value = 2 P(Z >2.688  )

From the z -table  

      P(Z > 2.688) = 0.0036

So  

      p-value = 2 *0.0036

        p-value = 0.0072

So from the p-value  obtained we see that p-value  <  \alpha so we reject the null hypothesis

Thus there is  sufficient evidence to conclude that the proportion of financial security is more fair in Feb 2010 is different from the proportion of financial security is more fair in Feb 2012.

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