Question

In: Statistics and Probability

The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0...

The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 2 3 4 Day shift 10 11 14 18 Afternoon shift 8 11 12 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shift? Hint: For the calculations, assume the day shift as the first sample. State the decision rule. (Round your answer to 2 decimal places.) Compute the value of the test statistic. (Round your answer to 3 decimal places.) What is the p-value? Between 0.025 and 0.05 Between 0.001 and 0.005 Between 0.005 and 0.01 What is your decision regarding H0? Reject H0 Do not reject H0

Solutions

Expert Solution

The null and alternative hypothesis is given as:

; the true mean of difference in the number of defective units produced in the Day shift and Afternoon shift is less than or equal to zero.

; the true mean of difference in the number of defective units produced in the Day shift and Afternoon shift is greater than zero. that is mean number of defective units produced on the Day Shift is greater than the Afternoon shift.

The formula for test statistic is:

;   with degrees of freedom, df=n-1

Day Day_Shift Afternoon_Shift d=Day_Shift-Afternoon_Shift
1 10 8 2
2 11 11 0
3 14 12 2
4 18 15 3

Standard deviation

Now the calculation for test statistic:

The test statistic is calculated as t=2.782

P-value:

Since it is a Right-tailed test, so the p-value is given as-

Or we can write that the p-value is between, , as 0.034 is between 0.025 and 0.05.

Decision:

Since,

In other words we say that, at the data does provide enough evidence to support the alternative hypothesis, i.e., , which means the true mean of difference of the number of defective units produced on Day shift and on the Afternoon shift is greater than zero.


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