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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 𝑿̅ = 𝟖𝟓 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 time slot effective?

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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 Yare independent, then Z=X-Ywill follow a normal distribution with mean \mu_X-\mu_Y and

variance \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 is \alpha=0.05,

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

The rejection region for this two-tailed test is 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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