Suppose a new standardized test is given to 100 randomly selected third-grade st

Question

Suppose a new standardized test is given to 100 randomly selected third-grade students in Arizona. The sample average score on the test is 60 points and the sample standard deviation is 10 points.

a. The authors plan to administer the test to all third-grade students in Arizona. Construct a 95% confidence interval for the mean score of all Arizona third graders.

b. Suppose the same test is given to 200 randomly selected third graders from Oregon, producing a sample average of 64 points and a sample standard deviation of 11 points. Construct a 90% confidence interval for the difference in mean scores between Arizona and Oregon.

c. State and test a null hypothesis that there is no difference in mean test performance between Arizona and Oregon third graders. Consider a significance level of 5% and indicate whether you are using a one or two sided test. Answer the following: What is the standard error of the difference in the two sample means? What is the p-value of the test of no difference in means versus some difference? Do you accept or reject the null hypothesis of no difference?

Explanation / Answer

(a) Sample size = 100

Sample mean x = 60 points

sample standard deviation s = 10 points

95% confidence interval for the mean score of all Arizona third graders = x +- t(99, 0.05)* s/ sqrt(n)

[Note: HEre we have took t- values. But as n >30 we can use Z value also]

= 60 +- 1.984 * 10/ sqrt(100)

= 60 +- 1.984

= (58.016, 61.984)

b. In oregon, Total sample size = 200

sample mean xO = 64

sample standard deivaiton so = 11

90% confidence interval for the mean score difference between Arizona and Oregon.

Total degrees of freedom = 298

pooled standard deviation sp = sqrt [(n1 -1)s12 + (n2 -1)s22 /n1 + n2 -2)] = sqrt [ (99 * 102 + 199 * 112 )/ (99 + 199) ] =  10.678

= (x Arizona -x oregon) +- t298,0.10 sp sqrt [1/n1 + 1/n2]

= (64 - 60) +- 1.65 * 10.678 * sqrt (1/100 + 1/200)

= 4 +- 2.158

= (1.842, 6.158)

C. Null HYpothesis : H0 : THere is no difference in mean test performance between Arizona and Oregon Third graders. Oregon = Arizona

ALternative Hypothesis : Ha : THere is significant difference in mean test performance between Arizona and Oregon THird graders. Oregon Arizona

As we will use the two sided test here because here it calls for difference only not for increase or decrease.

Standard error of the difference in two sample mean

se= sqrt [(n1 -1)s12 + (n2 -1)s22 /n1 + n2 -2)] * sqrt (1/n1 + 1/n2) = sqrt [ (99 * 102 + 199 * 112 )/ (99 + 199) ] * (1/100 + 1/200) ] = 1.3078

Test statistic

t = ((x Arizona -x oregon) / SE = ( 64 - 60)/ 1.3078 = 3.058

p - value = 2 * Pr (t > 3.058; 298) as n >>> 30 so we can find it from Z - table also

= 2 * 0.0012 = 0.0024 < 0.05 so statistically significant.

So, we will reject the null hypothesis and conclude that there is some difference in the two mean score.

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