Question

In: Statistics and Probability

all hypothesis testing problems must include the null and alternative hypotheses and report the p-value of...

all hypothesis testing problems must include the null and alternative hypotheses and report the p-value of the data.

A random sample of 45 students took an SAT preparation course prior to taking the SAT. The sample mean of their quantitative SAT scores was 575 with a s.d. of 90, and the sample mean of their verbal SAT scores was 530 with a s.d. of 110.
a) Construct 95% confidence intervals for the mean quantitative SAT and the mean verbal SAT scores of all students who take this course.
b) Construct 95% confidence intervals for the standard deviations of the QSAT and VSAT scores of all students who take this course.
c) What sample size would be needed to estimate the mean VSAT score with 95% confidence and with error of no more than 5 if it is assumed that the s.d. is no more than 110?
a) Suppose the mean scores for all students who took the SAT at that time was 535 for the quantitative and 505 for the verbal? Do the means for students who take this course differ from the means for all students at the 5% level of significance?

Solutions

Expert Solution

a)The 95% confidence interval for the mean quantitative SAT score is

where ,s = 90 , n=45

For 95% confidence with 44 df

therefore 95% confidence interval is

= ( 547.96 , 602.04)

The 95% confidence interval for the mean verbal SAT score is

where ,s = 110 , n=45

For 95% confidence with 44 df

therefore 95% confidence interval is

= ( 514.91 , 608.05)

b) 95% confidence interval for standard deviation of QSAT

s= 90, n=45, df=44

threfore 95% confidence interval for standard deviation of QSAT is

(74.51 , 113.69)

95% confidence interval for standard deviation of VSAT

s= 110, n=45, df=44

threfore 95% confidence interval for standard deviation of QSAT is

(91.06 , 138.95)

c)given

margin of error = 5

therefore n = 1859

sample size neede = 1859

d) QSAT

The null and alternative hypothesis

Test statistic

= 2.98

df =44

P value = 0.0047

Since P value < 0.05

we reject H0

At 5% level there is sufficient evidence to conclude that QSAT scores of students who take this course differ from means of all students .

VSAT

The null and alternative hypothesis

Test statistic

= 1.52

df =44

P value = 0.1357

Since P value > 0.05

we fail to reject H0

At 5% level there is not sufficient evidence to conclude that VSAT scores of students who take this course differ from means of all students .

orchestra answered 1 year ago

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