Answer :
Answer:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-1.337 -0}{\frac{7.755}{\sqrt{19}}}=-0.751[/tex]
[tex]p_v =P(t_{(18)}<-0.751) =0.2312[/tex]
So the p value is higher than the [tex] \apha=0.05[/tex] used, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0.
And the best option for this case would be:
Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.
Step-by-step explanation:
Previous concepts
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation
x=test value old , y = test value new
x: 55.8,51.7,76.6,47.5,48.6,11.4,30.6,53,21,58.5,42.6,61.2,26.8,11.4,56.3,46.1,72.8,42.2,51.3
y: 57.1,58.3,83.6,49.5,51.1,20.6,35.2,46.7,22.5,47.7,51.5,76.6,28.6,14.5,43.7,57,66.1,38.1,42.4
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_x- \mu_y \geq 0[/tex]
Alternative hypothesis: [tex]\mu_x -\mu_y <0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -1.3 -6.6 -7.0 -2.0 -2.5 -9.2 -4.6 6.3 -1.5 10.8 -8.9 -15.4 -1.8 -3.1 12.6 -10.9 6.7 4.1 8.9
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-1.337[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =7.755[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-1.337 -0}{\frac{7.755}{\sqrt{19}}}=-0.751[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=19-1=18[/tex]
Now we can calculate the p value, since we have a left tailed test the p value is given by:
[tex]p_v =P(t_{(18)}<-0.751) =0.2312[/tex]
So the p value is higher than the [tex] \apha=0.05[/tex] used, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0.
And the best option for this case would be:
Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.
Final answer:
To find the p-value, calculate the t-statistic and use the t-distribution. The p-value is 0.1207, so we do not reject the null hypothesis. There is not enough evidence to support the claim that the new online learning module increased placement scores.
Explanation:
To find the p-value, we need to calculate the t-statistic and then use the t-distribution to find the area to the left of the t-statistic.
- Calculate the differences between the new and old test scores for each student.
- Calculate the mean difference (the sample mean of the differences) and the standard deviation of the differences.
- Calculate the t-statistic using the formula: t = (mean difference - hypothesized mean difference) / (standard deviation / sqrt(sample size)).
- Find the p-value by looking up the t-statistic in the t-distribution table or using a calculator or software.
In this case, the p-value is 0.1207. Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis. Therefore, the correct decision is to Do not reject H0. The correct summary is that there is not enough evidence to support the claim that on average the new online learning module increased placement scores.
Learn more about Hypothesis testing here:
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