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Paha777 [63]
3 years ago
9

The two conditional relative frequency tables below show the results of a survey asking students whether they are taking a forei

gn language or not. A 4-column table with 3 rows. The first column has no label with entries middle school, high school, total. The second column is labeled taking a foreign language with entries 0.34, 0.66, 1.0. The third column is labeled not taking a foreign language with entries 0.64, 0.36, 1.0. The fourth column is labeled total with entries 0.4, 0.6, 1.0. Table B: Frequency of Foreign-Language Studies by Row A 4-column table with 3 rows. The first column has no label with entries middle school, high school, total. The second column is labeled taking a foreign language with entries 0.68, 0.88, 0.8. The third column is labeled not taking a foreign language with entries 0.32, 0.12, 0.2. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Which table could be used to answer the question "Assuming a student is taking a foreign language, what is the probability the student is also in high school?”
Mathematics
1 answer:
alex41 [277]3 years ago
8 0

Answer:

it's B

Step-by-step explanation:

Just took the test

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Answer:

t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

p_v =P(t_{(19)}>1.638)=0.0589    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

Step-by-step explanation:

Data given and notation    

\bar X=7.03 represent the sample mean

s=9.42 represent the sample standard deviation    

n=20 sample size    

\mu_o =3.58 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is higher than 3.58 :    

Null hypothesis:\mu \leq 3.58    

Alternative hypothesis:\mu > 3.58    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since is a right tailed test the p value would be:    

p_v =P(t_{(19)}>1.638)=0.0589    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

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