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2. Suppose a new standardized test is given to 100 randomly selected third-grade studentsin New Jersey. The sample average scoreYon the test is 58 points and the sample standarddeviationsYis 8 points.a. The authors plan to administer the test to all third-grade students in New Jersey.Construct a 95% con dence interval for the mean score of all New Jersey third graders.b. Suppose the same test is given to 200 randomly selected third graders from Iowa, pro-ducing a sample average of 62 points and sample standard deviation of 11 points. Constructa 90% con dence interval for the di erence in mean scores between Iowa and New Jersey.c. Can you conclude with a high degree of con dence that the population means for Iowaand New Jersey students are di erent? (What is the standard error of the di erence in thetwo sample means? What is thep-value of the test of no di erence in means versus somedi erence?)

Respuesta :

Answer:

a) n = 100

[tex]\bar{x} = 58[/tex]

s = 8

Confidence interval = 95%

Z at 95% = 1.96

Confidence interval formula : [tex]\bar{x}-z\frac{s}{\sqrt{n}}[/tex] to  [tex]\bar{x}-z\frac{s}{\sqrt{n}}[/tex]

Substitute the values :

Confidence interval : [tex]58-1.96 \times \frac{8}{\sqrt{100}}[/tex] to   [tex]58+1.96 \times \frac{8}{\sqrt{100}}[/tex]

Confidence interval : [tex]56.432[/tex] to   [tex]59.568[/tex]

b)n = 200

[tex]\bar{x} = 62[/tex]

s =11

Confidence interval = 90%

Z at 90% = 1.64

Confidence interval formula : [tex]\bar{x}-z\frac{s}{\sqrt{n}}[/tex] to  [tex]\bar{x}-z\frac{s}{\sqrt{n}}[/tex]

Substitute the values :

Confidence interval : [tex]62-1.64 \times \frac{11}{\sqrt{200}}[/tex] to   [tex]62+1.64 \times \frac{11}{\sqrt{200}}[/tex]

Confidence interval : [tex]60.724[/tex] to   [tex]63.275[/tex]

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