A student believes that the average grade across the entire university on the st

Question

A student believes that the average grade across the entire university on the statistics final examination was 87. A sample of 36 final examinations was taken from a current class. The average grade in the sample was 83.96 with a population standard deviation of 12. Please explain the steps. I’m unclear after stating the null and alternative hypotheses. I know we are use z-table

A. State the null and alternative hypotheses

B. Using the critical value approach, test the hypotheses at the 5% level of significance.

C. Using the p-value approach, test the hypotheses at the 5% level of significance.

Explanation / Answer

Step 1: State the hypothesis to be tested.
Null Hypothesis H0: = 0
Alternate Hypothesis H1: 0


Step 2: Determine a planning value for [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 - 0 )/( s/ SQRT(n) )

x-bar: Est. of the Pop. Mean (statistical mean of the sample) =
83.96
n: number of individuals in the sample =
36
s: sample standard deviation =
12
0: Population Mean =
87
significant digits =
3
Standardized Test Statistic t = ( 83.96 - 87 )/( 12 / SQRT( 36 )) =
1.52

Step 4: Using Students t distribution, 'lookup' the area outside of t = TDIST( 1.52 , 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 =
0.137
based on n -1 = 35 df (degrees of freedom).

Table look-up value shows area under the 35 df curve outside of t = +/- 1.52 is (approx.)
P value = 0.137 [2 * 0.0687] by addition of both 'tails' of t distribution.

Step 6: For P , fail to reject H0; and for P < , reject H0 with
0.95% confidence in the conclusion.

Conclusion: Null Hypothesis H0: = 0 (Null Hypothesis) is true with 95% confidence.

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