the question says....a manager believes that the average of a product is $75. A sample of 36 is taken. The average price is 72.50. The population variance is 144.
what is the null and alternative, using a critical value and P value, test the hypothesis at the 5% level of signifigance.
I thought you cant use both a critical value or p-value, its either or, right?|||ANSWER: Conclusion: Null Hypothesis H0: 渭 = 渭0 (Null Hypothesis) is true with 95% confidence.
Why??
SINGLE SAMPLE TEST, TWO-TAILED, 6 - Step Procedure for t Distributions, "two-tailed test"
Step 1: State the hypothesis to be tested.
Null Hypothesis H0: 渭 = 渭0
Alternate Hypothesis H1: 渭 鈮?渭0
Step 2: Determine a planning value for 伪 [level of significance] =
0.05
Step 3: From the sample data determine x-bar, s and n; then compute
Standardized Test Statistic: t = ( x-bar - 渭0 )/( s/ SQRT(n) )
x-bar: Est. of the Pop. Mean (statistical mean of the sample) =
72.5
n: number of individuals in the sample =
36
s: sample standard deviation =
24 [144/sqrt(36)]
渭0: Population Mean =
75
significant digits =
3
Standardized Test Statistic t = ( 72.5 - 75 )/( 24 / SQRT( 36 )) =
0.625
Step 4: Using Students t distribution, 'lookup' the area outside of t = TDIST( 0.625 , 35 , 2 ) using Excel TDIST(x, n-1 degrees_freedom, 2 tails)
using Excel TDIST(x, n-1 degrees_freedom, 2 tails)
Step 5: Area in Step 4 is equal to P value =
0.536
based on n -1 = 35 df (degrees of freedom).
Table look-up value shows area under the 35 df curve outside of t = +/- 0.625 is (approx.)
P value = 0.536 [2 * 0.268] by addition of both 'tails' of t distribution.
Step 6: For P 鈮?伪, fail to reject H0; and for P %26lt; 伪, reject H0 with
0.95% confidence in the conclusion.
Conclusion: Null Hypothesis H0: 渭 = 渭0 (Null Hypothesis) is true with 95% confidence.
Note: level of significance [伪] is the maximum level of risk an experimenter is willing
to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum
risk in making a Type I error).
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