Answer to Question #297436 in Statistics and Probability for lars

Question #297436

An experimental study was conducted by a researcher to determine if a new time slot has an effect on the performance of pupils in mathematics. Fifteen randomly selected learners participated in the study. Toward the end of the investigation, a standardized assessment was conducted. The sample mean was š‘æĢ… = šŸ–šŸ“ and the standard deviation s = 3. in the standardization of the test, the mean was 75 and the standard deviation was 10. based on the evidence at hand, is the new timeslot effective?


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Expert's answer
2022-02-14T16:26:00-0500

LetĀ "X"Ā have a normal distribution with meanĀ "\\mu_X"Ā and varianceĀ "\\sigma_X^2."

LetĀ "Y"Ā have a normal distribution with meanĀ "\\mu_Y"Ā and varianceĀ "\\sigma_Y^2."

IfĀ "X"Ā andĀ "Y"are independent, thenĀ "Z=X-Y"will follow a normal distribution with meanĀ "\\mu_X-\\mu_Y"Ā and

varianceĀ "\\sigma_X^2+\\sigma_Y^2."

"\\mu_{X-Y}=85-75=10"

"s_{X-Y}=\\sqrt{(3)^2+(10)^2}=\\sqrt{109}"

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, the sample standard deviation will be used.

Based on the information provided, the significance level isĀ "\\alpha=0.05,"

"df=n-1=15-1=14"Ā degrees of freedom, and the critical value for a two-tailed test isĀ "t_{0.025,14}= 2.145"

The rejection region for this two-tailed test isĀ "R=\\{t:|t|\\gt 2.145\\}"

The t-statistic is computed as follows:



"t=\\dfrac{\\mu_{X-Y}-\\mu}{s_{X-Y}\/\\sqrt{n}}=\\dfrac{10-0}{\\sqrt{109}\\over\\sqrt{15}}\\approx3.71"


Now, "t=3.71\\gt t_{0.025,14}=2.145" thus, we reject the null hypothesis and conclude that the data provides sufficient evidence to show that the new time slot is effective at 5% significance level.


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