Please need help with statistics problem #3 I dont know hoe to solve it. Please

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Please need help with statistics problem #3 I dont know hoe to solve it. Please show steps to take in order to solve. I will leave a great review.Thank you

Problem #3

a)A psychologist gives a verbal aptitude test to a random sample of 49 Oakland third graders. Assume that nationally on this test, third graders score on average 80 points with a standard deviation of 17. If Oakland third graders are like everyone else, what is the probability that the 49 Oakland third graders in the psychologist’s sample will have a mean score of 85 or higher?

b) On the verbal aptitude test, the 49 tested Oakland third graders actually attain an average score of 85 with a standard deviation of 17. Based on this result, which of the following two hypotheses would you have greater confidence to assert for the population of all Oakland third graders:

H0: Oakland third graders have verbal aptitudes scores, on average, equal to or lower than children nationally.

H1: Oakland third graders have verbal aptitudes scores, on average, higher than children nationally. Applying the principle of modus tollens, justify your answer.

c) Based on the results stated above, construct a 95% confidence interval for the population mean of all Oakland third graders on the verbal aptitude test.

Explanation / Answer

Mean ( u ) =80
Standard Deviation ( sd )=17
Number ( n ) = 49
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a.
P(X < 85) = (85-80)/17/ Sqrt ( 49 )
= 5/2.4286= 2.0588
= P ( Z <2.0588) From Standard NOrmal Table
= 0.9802                  
P(X > = 85) = 1 - P(X < 85)
= 1 - 0.9802 = 0.0198                  

b.
H0: Oakland third graders have verbal aptitudes scores, on average, equal to or lower than children nationally

c.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=85
Standard deviation( sd )=17
Sample Size(n)=49
Confidence Interval = [ 85 ± t a/2 ( 17/ Sqrt ( 49) ) ]
= [ 85 - 2.011 * (2.429) , 85 + 2.011 * (2.429) ]
= [ 80.116,89.884 ]

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