A random sample of 11 nursing students from Group 1 resulted in a mean score of 65 with a standard deviation of 4.7. A random sample of 16 nursing students from Group 2 resulted in a mean score of 74 with a standard deviation of 2.5. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2

The calculated value of t= -1.118860 does not lie in the critical region t= -1.714 or t= -2.5. Therefore we accept our null hypothesis that the mean score for Group 1 is significantly lower than the mean score for Group 2 for both significance level of 0.05 and 0.01

Step-by-step explanation:

Groups               Sample Size      Mean        Standard Deviation

1                              11                      65                   4.7

2                            16                      74                    2.5

Formulate null and alternate hypotheses as

H0 : u1 < u2 against Ha: u1 ≥ u 2

That is the mean of the group 1 is lower than the mean of the group 2

Degrees of freedom is calculated df = υ= n1+n2- 2= 11+16-2= 23

The significance level alpha is chosen to be ∝ = 0.05

The critical region t ≤ -t (0.05, 23) = -1.714

And if we choose alpha = 0.01 the critical region would be  

t ≤ -t (0.05, 23) = -2.5

Here the difference between the sample means is x`1- x`2= 65-74= -9

The pooled estimate for the common variance σ² is

Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]

= 1/11+16-2 [ (65)²+(74)²]

= 1/23[ 4225+5476]

= 1/23[9701]

=421.782

Sp =√ 421.782=20.5373

The test statistic is

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t= -9/ 20.5373√1/11+ 1/16

t= -9/8.0439

t= -1.118860

The calculated value of t= -1.118860 does not lie in the critical region t= -1.714 or t= -2.5. Therefore we accept our null hypothesis that the mean score for Group 1 is significantly lower than the mean score for Group 2 for both significance level of 0.05 and 0.01

6e14 is the answer.