<span>f(x)= 50x
</span><span>y = k x
k is variation
so k= 50
D is the right option
hope it helps
</span>
The conclusion is that since the value of 0.028 is less than 0.05 so we reject the null hypothesis at 5% level of significance and conclude that the mean of the difference in SAT writing scores for all students who take SAT prep class is equal to 0.
<h3>What is a null hypothesis?</h3>
A null hypothesis is a hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling error.
Here, the effect of a one-day SAT prep class at a 5% level of significance was assessed. Since the value of 0.028 is less than 0.05 so we reject the null hypothesis at 5% level of significance and conclude that the mean of the difference in SAT writing scores for all students who take SAT prep class is equal to 0.
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