Question #336799

APPLICATION:





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 x̄= 75 and s=5. In standardization of the test, the mean was





65 and standard deviation was 8. Based on the evidence at





hand, is the new time slot effective? Use α = 0.05.




1
Expert's answer
2022-05-05T16:53:09-0400

Difference of two independent normal variables

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=7565=10\mu_{X-Y}=75-65=10sXY=(5)2+(8)2=89s_{X-Y}=\sqrt{(5)^2+(8)^2}=\sqrt{89}

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=0.05,\alpha=0.05,

df=n1=151=14df=n-1=15-1=14 degrees of freedom, and the critical value for a two-tailed test is tc=2.144787.t_c=2.144787.

The rejection region for this two-tailed test is R={t:t>2.144787}R=\{t:|t|>2.144787\}

The t-statistic is computed as follows:


t=μXYμsXY/n=10089/154.105354t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{10-0}{\sqrt{89}/\sqrt{15}}\approx4.105354

Since it is observed that t=4.105354>2.144787=tc,|t|=4.105354>2.144787=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed α=0.05,\alpha=0.05, df=14,t=4.105354df=14, t=4.105354 is p=0.001071,p=0.001071, and since p=0.001071<0.05=α,p=0.001071<0.05=\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=0.05\alpha=0.05 significance level.

Therefore, there is enough evidence to claim that the new time slot can be effective, at the α=0.05\alpha=0.05 significance level.


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