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il63 [147K]
4 years ago
15

Conscientiousness is a tendency to show self-discipline, act dutifully, and aim for achievement. The trait shows a preference fo

r planned rather than spontaneous behavior. A random sample of 650 students is asked to fill out the Hogan Personality Inventory (HPI) to measure their level of conscientiousness. The 350 graduate students had a mean score of 153 with a standard deviation of 21. The 300 undergraduate students scored an average of 148 with a standard deviation of 16. We are interested in determining whether graduate students score higher, on average, on the HPI than undergraduate students. What is the value of the test statistic for testing these hypotheses using a pooled two sample t-test?
Mathematics
1 answer:
MAVERICK [17]4 years ago
6 0

Answer:

a

The Null hypothesis represented as

     H_0: \mu_1 -  \mu_2  = 0

   

The Alternative hypothesis represented as

     H_a: \mu_1 -  \mu_2  <  0

b

p-value =  P(Z <  -3.37  )  = 0.000376

c

There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

Step-by-step explanation:

From the question we are told that

   The population size is  n= 650

    The  sample size for graduates is  n_1 =  300

     The sample  mean for graduates is \= x _1 =  148

      The sample  standard deviation for graduates is \sigma_1  =  16

    The  sample size for under-graduates is n _2 = 350

         The sample  mean for under-graduates is \= x _2 =  153

         The sample  standard deviation for graduates is \sigma_2  =  21

The Null hypothesis represented as

     H_0: \mu_1 -  \mu_2  = 0

   

The Alternative hypothesis represented as

     H_a: \mu_1 -  \mu_2  <  0

Where \mu_1 \ and \  \mu_2 are the population mean

  Now the test statistic is mathematically represented as

          t =  \frac{(\= x_1 - \= x_2 ) }{ \sqrt{ \frac{ (n_1 - 1 )\sigma_1 ^2 + (n_2 - 1)\sigma_2^2}{n_1 +n_2 -2} }  * \sqrt{ \frac{1}{n_1}  + \frac{1}{n_2} } }

substituting values

       t =  \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} }  * \sqrt{ \frac{1}{300}  + \frac{1}{350} } }

      t = -3.37

The p-values is mathematically evaluated as

     p-value =  P(Z <  -3.37  )  = 0.000376

The above answer is gotten using a p-value calculator  at (0.05) level of significance

     Looking the p-value we see that it is less than the level of significance (0.05)  so Null hypothesis is rejected

Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

   

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