In: Statistics and Probability
Test the given claim. Assume that a simple random sample is
selected from a normally distributed population. Use either
the P-value method or the traditional method of testing hypotheses.
Company A uses a new production method to manufacture aircraft
altimeters. A simple random sample of new altimeters resulted in
errors listed below. Use a 0.05 level of significance to test the
claim that the new production method has errors with a standard
deviation greater than 32.2 ft, which was the standard deviation
for the old production method. If it appears that the standard
deviation is greater,
does the new production method appear to be better or worse than
the old method? Should the company take any action?
− 41, 75, − 23, − 71, − 44, 14, 19, 52, − 5, − 54, − 108, −
108
What are the null and alternative hypotheses?
A. H0: σ = 32.2 ft
H1: σ < 32.2 ft
B. H0: σ < 32.2 ft
H1: σ = 32.2 ft
C. H0: σ = 32.2 ft
H1: σ ≠ 32.2 ft
D. H0: σ ≠ 32.2 ft
H1: σ = 32.2 ft
E. H0: σ = 32.2 ft
H1: σ > 32.2 ft
F. H0: σ > 32.2 ft
H1: σ = 32.2 ft
Find the test statistic.
χ2 = (Round to two decimal places as needed.)
Determine the critical value(s).
The critical value(s) is/are ---------(Use a comma to separate
answers as needed. Round to two decimal places as needed.)
Since the test statistic is (1) ------------the critical
value(s), (2)-----------Ho . There is (3)---------- evidence to
support the claim that the new production method has errors with a
standard deviation
greater than 32.2 ft.
The variation appears to be (4)--------- than in the
past, so the new method appears to be (5)---------- , because there
will be (6)-------- altimeters that have errors. Therefore, the
company (7)---------- take immediate action to reduce the
variation.
(1) between
equal to
less than
greater than
(2)fail to reject
reject
(3) sufficient
insufficient
(4) about the same
less
greater
(5) similar
better
worse
(6) the same number of
more
fewer
(7) should not
should
Answers
The null hypothesis is H0: = 32.2 ft. and
the alternate hypothesis is H1: > 32.2 ft.
Option E is the correct option
This is a one - sided right - tailed test
The following table shows the calculations -
Errors(x) |
(x - ) |
(x - )^2 |
|
-41 |
-16.5 |
272.25 |
|
75 |
99.5 |
9900.25 |
|
-23 |
1.5 |
2.25 |
|
-71 |
-46.5 |
2162.25 |
|
-44 |
-19.5 |
380.25 |
|
14 |
38.5 |
1482.25 |
|
19 |
43.5 |
1892.25 |
|
52 |
76.5 |
5852.25 |
|
-5 |
19.5 |
380.25 |
|
-54 |
-29.5 |
870.25 |
|
-108 |
-83.5 |
6972.25 |
|
-108 |
-83.5 |
6972.25 |
|
Total |
-294 |
0 |
37139 |
Number of observations, n = 12
Mean, = -294/12 = -24.5
Sample Standard Deviation, s’ ={ (x - )^2 / (n - 1)}^0.5 = 58.1057
The test statistic for this test is 2 = (n - 1)(s'^2) / (^2) which follows Chi - Square distribution with (n - 1) degrees of freedm under the null hypothesis
Substituting all values,
2 = 35.82
The Significance level = 5% = 0.05
and the degrees of freedom = 12 - 1 = 11
The critical value is 19.67
Generally if the test statisic is greater than the critical value we reject the null else we fail to reject the null hypothesis at level
Since the test statistic is greater than (1) the critical value, reject (2) H0. There is sufficient (3) evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft.
The variation appears to be greater (4) than in the past, so the new method appears to be worse (5), because there will be more (6) altimeters that have errors. Therefore, the company should (7) take immediate action to reduce the variation.