The students' attitudes were measured by administering the computer anxiety rating scale (cars). a random sample of 11 nursing students from group 1 resulted in a mean score of 63.3 with a standard deviation of 3.7. a random sample of 13 nursing students from group 2 resulted in a mean score of 70.2 with a standard deviation of 6.6. can you conclude that the mean score for group 1 is significantly lower than the mean score for group 2? let μ1 represent the mean score for group 1 and μ2 represent the mean score for group 2. use a significance level of α=0.05 for the test. assume that the population variances are equal and that the two populations are normally distributed.

There is enough evidence to claim that population mean 1 is less than population 2  at the 0.05 significance level.

Step-by-step explanation:

We are given the following in the question:

Group 1:

Group 2:

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

Since, the population variances are equal and that the two populations are normally distributed, we use t-test(pooled test) for difference of two means.

Formula:

Pooled standard deviation

Putting all the values we get:

Now,

Since,

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

We conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2.

There is enough evidence to claim that population mean 1 is less than population 2  at the 0.05 significance level.

plz yuo

step-by-step explanation:

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2