Question #338016

A teacher conducted a study to know if blended learning affects the students’ performances. A class of 30 students of Grade 11 was surveyed and found out that their mean score was 83 with a standard deviation of 4. A study from other country revealed that with a standard deviation of 3. Test the hypothesis at 0.10 level of significance.

Parameter: _______________________________________

Statistic: ________________________________

Null Hypothesis: ____________________________________________________

Alternative hypothesis: _____________________________________________


1
Expert's answer
2022-05-08T14:14:48-0400

Parameter: Difference of two independent normal variables μXY\mu_{X-Y}

Let XX have a normal distribution with mean μX\mu_X and variance σX2.\sigma_X^2.

Let YY have a normal distribution with mean μY\mu_Y and variance σY2.\sigma_Y^2.

If XX and YYare independent, then XYX-Ywill follow a normal distribution with mean μXμY\mu_X-\mu_Y and variance σX2+σY2.\sigma_X^2+\sigma_Y^2.

μXY=8380=3\mu_{X-Y}=83-80=3sXY=(3)2+(4)2=5s_{X-Y}=\sqrt{(3)^2+(4)^2}=5



Statistic: tt- statistic


The following null and alternative hypotheses need to be tested:

H0:μ=0H_0:\mu=0

H1:μ0H_1:\mu\not=0

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.10,\alpha=0.10,

df=n1=301=29df=n-1=30-1=29 degrees of freedom, and the critical value for a two-tailed test is tc=1.699127.t_c=1.699127.

The rejection region for this two-tailed test is R={t:t>1.699127}R=\{t:|t|>1.699127\}

The t-statistic is computed as follows:


t=μXYμsXY/n=305/303.2863t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{3-0}{5/\sqrt{30}}\approx3.2863

Since it is observed that t=3.2863>1.699127=tc,|t|=3.2863>1.699127=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed α=0.10,\alpha=0.10, df=29,t=3.2863df=29, t=3.2863 is p=0.002659,p=0.002659, and since p=0.002659<0.10=α,p=0.002659<0.10=\alpha, it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is different than 0, at the α=0.10\alpha=0.10 significance level.

Therefore, there is enough evidence to claim that blended leaming affects the students' performances, at the α=0.10\alpha=0.10 significance level.


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