Answer to Question #338019 in Statistics and Probability for KLTino

Question #338019

Determine the given and compute the appropriate test statistic of the problem below.

Construct the rejection region of the problem below


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.



1
Expert's answer
2022-05-10T14:40:30-0400

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

Let "X" have a normal distribution with mean "\\mu_X" and variance "\\sigma_X^2."

Let "Y" have a normal distribution with mean "\\mu_Y" and variance "\\sigma_Y^2."

If "X" and "Y"are independent, then "X-Y"will follow a normal distribution with mean "\\mu_X-\\mu_Y" and variance "\\sigma_X^2+\\sigma_Y^2."

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



Statistic: "t-" statistic


The following null and alternative hypotheses need to be tested:

"H_0:\\mu=0"

"H_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 "\\alpha=0.10,"

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

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

The t-statistic is computed as follows:


"t=\\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=t_c," it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed "\\alpha=0.10," "df=29, t=3.2863" is "p=0.002659," and since "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 "\\mu" is different than 0, at the "\\alpha=0.10" significance level.

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


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