(27)
(i) Correct procedure to use: 1 Sample t test
(ii) Parameter of interest: Quantitative NAAL score for young adults
(iii) H0: Null Hypothesis: 290
HA: Alternative Hypothesis: 290
(iv) Level of significance = = 0.05
(28)
(a)
Conditions for the test:
(i) The dependent variable is continuous
(ii) The observations are independent of each other
(iii) The dependent variable is approximately normally distributed
(iv) The dependent variable does not contain any outliers
(b)
How these conditions are met: Explained in the description of the problem.
(c)
SE = s/
= 112/ = 6.2222
Test statistic is:
t = (276 - 290)/6.2222 = - 2.25
(d)
t score = - 2.25
ndf = 324 - 1 = 323
One Tail - Left Side Test
By Technology, P - Value = 0.0126
(29)
Since P - Value is less than , the difference is significant. Reject null hypothesis.
Conclusion:
The data support the claim that mean quantitative NAAL score for
all young adults is less than 290.
(30)
95% Confidence Interval: (263.98, 288.02) defines a range of values that we can be 95% certain contains the population mean.
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