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It is reported that scores on a particular test of historical trivia given to high school students are approximately normally distributed with a mean of 85. Mrs. Rose believes that her high school seniors will score significantly better than the national average on this test based on the instruction they receive in her American History course. At the end of the semester, Mrs. Rose administers the historical trivia test to her students. The students score an average of 89 on this test, with a standard deviation of 2. After conducting the appropriate statistical test, Mrs. Rose finds that the p-value is .0025. Unfortunately, Mrs. Rose knows little about statistics and cannot understand what this p-value is telling her.
1. Interpret the p-value for Mrs. Rose.

Respuesta :

Answer:

If we compare the p value obtained and a significance level assumed for example [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the high school seniors from Mrs Rose have scores higher than 85 at 1% of significance.  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=89[/tex] represent the mean average  

[tex]s=2[/tex] represent the sample standard deviation for the sample  

[tex]n[/tex] sample size  

[tex]\mu_o =85[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean scores are higher than 85 the national average, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 85[/tex]  

Alternative hypothesis:[tex]\mu >85[/tex]  

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:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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".  

3) Calculate the statistic  

We can replace in formula (1) , but since we don't have the sample size we can't calculate it.

4)P-value  

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

[tex]p_v =P(z>z_{calculated})=0.0025[/tex]

That's the information given by the problem.  

5) Conclusion  

If we compare the p value obtained and a significance level assumed for example [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the high school seniors from Mrs Rose scores higher than 85 at 1% of significance.  

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