Answer :
Answer:
a
The Null hypothesis represented as
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
The Alternative hypothesis represented as
[tex]H_a: \mu_1 - \mu_2 < 0[/tex]
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 [tex]n= 650[/tex]
The sample size for graduates is [tex]n_1 = 300[/tex]
The sample mean for graduates is [tex]\= x _1 = 148[/tex]
The sample standard deviation for graduates is [tex]\sigma_1 = 16[/tex]
The sample size for under-graduates is [tex]n _2 = 350[/tex]
The sample mean for under-graduates is [tex]\= x _2 = 153[/tex]
The sample standard deviation for graduates is [tex]\sigma_2 = 21[/tex]
The Null hypothesis represented as
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
The Alternative hypothesis represented as
[tex]H_a: \mu_1 - \mu_2 < 0[/tex]
Where [tex]\mu_1 \ and \ \mu_2[/tex] are the population mean
Now the test statistic is mathematically represented as
[tex]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} } }[/tex]
substituting values
[tex]t = \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} } * \sqrt{ \frac{1}{300} + \frac{1}{350} } }[/tex]
[tex]t = -3.37[/tex]
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
The answer is "3.37".
Following are the two-Sample of the T-Test and CI:
[tex]\boxed{\boxed{\begin{matrix}Sample& N &Mean& StDev& SEMean\\ 1 &350 &153.0& 21.0 &1.1 \\ 2& 300& 148.0& 16.0& 0.92\end{matrix}}}[/tex]
Calculating the difference:
[tex]= \mu (1) - \mu (2)\\\\[/tex]
The estimated difference: [tex]5.00[/tex]
When the [tex]95\%[/tex] of CI so, the difference:[tex](2.09, 7.91)[/tex]
Therefore the T-Test difference value = 0
[tex]\to (vs \neq): T-Value = 3.37 \\\\\to P-Value = 0.001 \\\\\to DF = 648[/tex]
When the both use Pooled so, the StDev = 18.8583
Therefore, the final value of t is "3.37".
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