Question

1.For each situation, state the null and alternative hypotheses: (Type "mu" for the symbol ?μ , e.g.  mu >> 1 for the mean is greater than 1,  mu << 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1. Please do not include units such as "mm" or "$" in your answer.)

Harry thinks that prices in Caldwell are higher than the rest of the country. He reads that the nationwide average price of a certain brand of laundry detergent is $19.95 with standard deviation $2.18. He takes a sample from 4 local Caldwell stores and finds the average price for this same brand of detergent is $22.13.

?0 :

?? :

2.A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below.

Person	A	B	C	D	E	F	G	H	I	J
Test Score	84	80	91	96	88	79	74	67	94	70

Use a 0.010.01 significance level to test the claim that people do better with the new edition. Assume the standard deviation is 10.5. (Note: You may wish to use statistical software.)

Construct a 9999% confidence interval for the mean score for students using the new text.
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Answer 1

1)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0:mu = 19.95
Alternative Hypothesis, Ha: mu > 19.95

2)

sample mean, xbar = 82.3
sample standard deviation, σ = 10.5
sample size, n = 10

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 10.5/sqrt(10)
ME = 8.57

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (82.3 - 2.58 * 10.5/sqrt(10) , 82.3 + 2.58 * 10.5/sqrt(10))
CI = (73.7334 , 90.8666)