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Veseljchak [2.6K]
2 years ago
7

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60

0 to 2400.
Suppose that PrepIt! is a company that offers classes to help students prepare for the SAT exam. In their ad, PrepIt! claims to produce "statistically significant" increases in SAT scores. This claim comes from a study in which 427 PrepIt! students took the SAT before and after PrepIt! classes. These students are compared to 2,733 students who took the SAT twice, without any type of formal preparation between tries.

We also conduct a hypothesis test with this data and find that students who retake the SAT without PrepIt! also do significantly better (p-value < 0.0001). So now we want to determine if PrepIt! students improve more than students who retake the SAT without going through the PrepIt! program. In a hypothesis test, the difference in sample mean improvement ("PrepIt! gain" minus "control gain") gives a p-value of 0.004. A 90% confidence interval based on this sample difference is 3.0 to 13.0.

What can we conclude?

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.
Mathematics
1 answer:
kodGreya [7K]2 years ago
5 0

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

n_1 represent the sample after the preparation

n_2 represent the sample without preparation  

\bar x_1 =678 represent the mean sample after preparation

\bar x_2 =1837 represent the mean sample without preparation

s_1 =197 represent the sample deviation after preparation

s_2 =328 represent the sample deviation without preparation

\alpha=0.1 represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})} (1)  

The point of estimate for \mu_1 -\mu_2

The appropiate degrees of freedom are df=n_1+ n_2 -2

Since the Confidence is 0.90 or 90%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}  

After replace in the formula for the confidence interval we got this:

3.0 < \mu_1 -\mu_2

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

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Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......

where, n = number of trials (samples) taken = 20 children

            r = number of success = none of the 20 children

            p = probability of success which in our case is probability that

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<u><em>Let X = Number of children that are unvaccinated</em></u>

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

           P(X = 0)  =  \binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}

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2 years ago
What is 15 tens - 1 gross ​
vampirchik [111]

Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

what is 15 tens - 1 gross

we know that

15 tens means ----> That you are adding 10, 15 times or multiplying 10 by 15, which gives you

10(15)=150

1 gross means ---> That you are adding 10, 12 times or multiplying 10 by 12

which gives you

10(12)=120

so

The algebraic expression of 15 tens - 1 gross is equal to

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Convert to word expression

3  tens

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Answer:

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Step-by-step explanation:

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75 % of 44 = 75/100 * 44= 0.75 *44 = 33

So the third quartile lies at 33.

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