The null and alternate hypotheses are: H_0: mu_d lessthanorequalto 0 H_1: mu_d >

Question

The null and alternate hypotheses are: H_0: mu_d lessthanorequalto 0 H_1: mu_d > 0 The following sample information shoe's the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. At the .01 significance level, can we conclude there are more defects produced on the day shift? State the decision rule. Reject H_0 if t > Compute the value of the test statistic. Value of the test statistic What is the p-value? p-value What is your decision regarding H_0?

Explanation / Answer

1. Degrees of freedom = n - 1 = 4 - 1 = 3

For 3 degrees of freedom and a right tailed test,

Critical t value = 4.54

So,

Reject Ho if t > 4.54

The statistical software output for this problem is:

Paired T hypothesis test:
D = 1 - 2 : Mean of the difference between Day Shift and Afternoon Shift
H0 : D = 0
HA : D > 0
Hypothesis test results:

Hence,

2. Test statistic = 5.745

3. p - value = 0.0052

4. Reject Ho

Difference Mean Std. Err. DF T-Stat P-value Day Shift - Afternoon Shift 2.75 0.47871355 3 5.7445626 0.0052

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