Number of defective monitors manufactured in day shift and afternoon shift is to be compared. A sample of the production from six day shifts and eight afternoon shifts revealed the following number of defects.
Day 4 5 8 6 7 9
Afternoon 9 8 10 7 6 14 11 5
Is there a difference in the mean number of defects per shift? Choose an appropriate significance level.
(a) State the null hypothesis and the alternative hypothesis.
(b) What is the decision rule?
(c) What is the value of the test statistic?
(d) What is your decision regarding the null hypothesis?
(e) What is the p-value?
(f ) Interpret the result.
(g) What assumptions are necessary for this test?
(Typed answer preferred)
a)
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
b)
α=0.05
Degree of freedom, DF= n1+n2-2 =
12
t-critical value , t* = ± 2.179 (excel formula
=t.inv(α/2,df)
reject Ho: if test stat <-2.179 or t > 2.179
c)
Sample #1 ----> 1
mean of sample 1, x̅1= 6.500
standard deviation of sample 1, s1 =
1.871
size of sample 1, n1= 6
Sample #2 ----> 2
mean of sample 2, x̅2= 8.750
standard deviation of sample 2, s2 =
2.915
size of sample 2, n2= 8
difference in sample means = x̅1-x̅2 =
6.5000 - 8.8 =
-2.250
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 2.5331
std error , SE = Sp*√(1/n1+1/n2) =
1.3680
t-statistic = ((x̅1-x̅2)-µd)/SE = (
-2.2500 - 0 ) /
1.37 = -1.6447
d)
Decision: t-stat >-2.179 , so, Do not
Reject Ho
e)
p-value = 0.1260
(excel function: =T.DIST.2T(t stat,df) )
f)
There is not enough evidence that there is a difference in the mean number of defects per shift
g)
sample are random and independent
population from which samples are taken should be normally distributed
Number of defective monitors manufactured in day shift and afternoon shift is to be compared. A...
The null and alternative hypotheses are: H H :Hd < 0 :Hd > 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 Day shift Afternoon shift 1 10 8 2 12 9 3 15 12 4 11 15 At the 0.05 significance level, can we conclude there are more defects produced on the Afternoon shift? a. State the decision rule....
The null and alternate hypotheses are: HO: Haso H:Ha> 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 shift Afternoon shift 11 10 10 12 14 12 17 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shit? Hint: For the calculations, assume the day shift as the first sample. a. State the...
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 11 12 14 18 Afternoon shift 9 10 13 16 At the .005 significance level, can we conclude there are more defects produced on the afternoon shift? Hint: For the...
The null and alternate hypotheses are: Ho HdsO 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 Day shift Afternoon shift 89 12 17 10 12 15 18 At the 0.025 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. a. State the decision...
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 11 10 14 19 Afternoon shift 10 9 14 16 At the .01 significance level, can we conclude there are more...
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 12 15 19 Afternoon shift 8 11 12 20 At the 0.050 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...
Compute the value of the test statistic The null and alternate hypotheses are: Ho Maso HP>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 2 12 10 1 10 9 Day shift Afternoon shift 13 14 4 18 15 At the 0.050 significance level, can we conclude there are more defects produced on the day shift? Hint: For the calculations,...
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the percentage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. This hypothesis test is a two-tailed test about a proportion. True...
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 10 16 17 Afternoon shift 9 10 14 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...
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 12 12 16 19 Afternoon shift 10 10 12 15 At the .05 significance level, can we conclude there are more...