Mathematics, 15.01.2020 21:31 ajmitchell04

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.

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Answer from: ryandelanomartinez

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:

$\mu_1 = 63.3\\\sigma_1 = 3.7\\n_1 = 11$

Group 2:

$\mu_2 = 70.2\\\sigma_2 = 6.6\\n_2 = 13$

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

$H_{0}: \mu_1 \geq \mu_2\\H_A: \mu_1 < \mu_2$

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

$s_p = \sqrt{\displaystyle\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2 }{n_1 + n_2 - 2}}$

$t_{stat} = \displaystyle\frac{\mu_1-\mu_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\text{Degree of freedom} = n_1 + n_2 - 2$

Putting all the values we get:

$s_p = \sqrt{\displaystyle\frac{(11-1)(3.7)^2 + (13-1)(6.6)^2 }{11 + 13 - 2}} = \sqrt{29.9827} = 5.48$

$t_{stat} = \displaystyle\frac{63.3-70.2}{5.48\sqrt{\frac{1}{11}+\frac{1}{13}}} = -3.073$

$\text{Degree of freedom} = 11 + 13 - 2 = 22$

Now,

$t_{critical} \text{ at 0.05 level of significance, 22 degree of freedom } = -1.717$

Since,

$t_{stat} < t_{critical}$

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.

Answer from: Quest

Find the perimeter of the figure shown above with a side length 5 units.(each figure is a square).a.10 unitsc.30 unitsb.20 unitsd.40 units select the best answer from the choices providedabcdmark this and return

Answer from: Quest

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2 polynomial long division

dividing : x4+5x3-4x2-8x+6

("dividend")

by : x-1 ("divisor")

dividend x4 + 5x3 - 4x2 - 8x + 6

- divisor * x3 x4 - x3

remainder 6x3 - 4x2 - 8x + 6

- divisor * 6x2 6x3 - 6x2