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§ Examples- Lesson 2 §
Example 1
A magazine article claims that the average life time of batteries when used in ordinary toys is not the same as 10 years ago; it claims the average life time of batteries (m) is now 250 minutes. Your firm sells battery powered toys, so the value of this population mean is of vital interest to you. To investigate the article's claim, you randomly select 36 batteries for testing. Use Excel to solve this problem. Your results are:
Given:
Data: 240, 250, 230, 245, 235, 241, 239, 242, 238, 243, 237, 244, 236, 246, 234, 248, 232, 249, 231, 250, 230, 230, 250, 210, 270, 211, 269, 200, 280, 199, 281, 195, 285, 190, 290,240
n = 36
xbar = 240
s = 24
The Five-Step Procedure for Hypothesis Testing-
(1) Set up the Null Hypothesis, Ho, and Alternative Hypothesis, Ha.
Ho: m = 250; Ha: m ¹ 250
(click me)
(a) The number to the right of the equal sign is always the claimed parameter value (never xbar ). This is true for both Ho and Ha.
(b) The signs used in Ho and Ha are always the mathematical opposites of each other. Some form of an "equal sign" ( = ) is always associated with the Ho (never Ha).
(c) Question: Is our statistic, xbar = 240,"close enough" to the claimed parameter value, m = 250, to FTR (support) Ho?
[1] We do not expect our point estimate, xbar = 240, to be exactly equal to the claimed parameter value, m = 250.
[2] Is xbar "statistically close enough" to m to believe the Ho? Use the standard error, sxbar or s xbar, and Z or t score to measure the statistical distance between them.
What is the probability that the sample xbar in a problem would be equal to
the population m? 1.00
zero
0.50
(click one)
(2) Define the test statistic. Use Z or t?
(a)
Source of Standard Deviation
|
|
|
Population s |
Sample s |
|
Sample Size |
Small df £ 30 |
Z |
t |
|
|
Large df > 30 |
Z |
Z or t |
(b) Use
z to calculate the size of the difference between
xbar = 240 and m
= 250.
(click me)
Would it be possible to use t in this problem? Yes
No
(click one)
(3) Define a region(s) of rejection based on a.
(a) What
a can be tolerated?
What percent of the time will you reject Ho
when Ho is true?
(click me)
(b) Use a = 0.05.
(c) For a two - tail test, each tail will have a / 2 in it.
[1] a/2 = 0.025
[2] 0.5 - a/2 = 0.5 - 0.025 =0.4750 (look up in Z table)
(d) table value of test statistic ± Z0.025 = ± 1.96 (critical value)
[1] The table value is always associated with a (never with xbar), is found in a Z (or t ) table with subscript a or a/2, and defines the region(s) of rejection.
What would be the table Z (critical value) in this problem if alpha were equal
to 0.10? 2.575
1.645
1.282
(click one)
(4) Calculate the value of the test statistic and carry out the test
(a) The calculated (or computed ) value of the test statistic is always associated with xbar (never a), has a "star" next to it ( Z* or t*), and is calculated by you (not found in a table).
(b) Z* = [xbar - mo ] / [s /Ö n]
= [ 240 - 250 ] / [ 24/ Ö 36 ]
= [ - 10] / [24 / 6]
= [- 10]
/ [4.0] = - 2.50 (click
me)
(c) Note that "s" is used here in the calculation of Z*, because the sample size is large, n = 36.]
(d) and carry out the test:
Large-Sample Two-Tailed Tests on the Population Mean
Ho: µ = µo
Ha: µ ¹ µo
Reject Ho if |Z*| > Z a /2
FTR (support) Ho if |Z*| £ Z a /2
(e) Z* = - 2.50
|Z*| = 2.50
Z0.025 = 1.96
[1] |Z*| > Z a /2
[2] 2.50
> 1.96, therefore Reject Ho
(click me)
What would be the conclusion to this problem if alpha were equal to 0.01? FRT(Support)
Ho Reject
Ho (click one)
(5) Give a conclusion in terms of the original problem or question.
(a) Since at a = 0.05 the sample xbar = 240 is statistically so much smaller than the hypothesized parameter of 250, one can not believe that Ho: m = 250 is true; therefore,
(b) Reject Ho. (|Z*| > Z0.025 i.e. 2.50 > 1.96 ). There is no statistically significance evidence that the mean life time of the batteries is equal to 250.
"There is statistically significance evidence that the mean life time of
the batteries is not equal to 250." Is this statement consistent with the
conclusion in (5)-(b) above. Yes
No
(click one)
(6) Diagram if a = 0.05:
(a) Ho: µ = 250; Ha: µ ¹ 250
(b) a = 0.05 a /2 = 0.025 0.5 - a /2 = 0.5 - 0.0250 = 0.4750 Z table
Z0.025 = ±1.96
(c) Z* = [xbar - m ] / [s /Ö n] = [ 240 - 250 ] / [ 4.0 ] = - 2.50
(d) |Z*| > Z a /2
2.50 > 1.96, therefore Reject Ho.
(e) Since at = 0.05 the sample xbar = 240 is statistically so much smaller than the hypothesized parameter of 250, one can not believe that Ho: m = 250 is true; therefore,
(f) Reject Ho. (|Z*| > Z0.025 i.e. 2.50 > 1.96). There is no statistically significant evidence that the mean life time of the batteries is equal to 250.
Example 2
A 95% confidence interval for the mean time that it takes to commute to work in Phoenix is 1.1 hours to 1.5 hours. The time it takes to commute to work in Phoenix is normally distributed. Is there sufficient evidence to indicate that the mean time to complete the route is different from 1.0 hour? Use a 5% significance level.
(A) Ho: µ = 1; Ha: µ ¹ 1
(B) The not equal "¹ " sign indicates a two - tailed hypothesis test, which is necessary when using a CI to do the test.
(C) Since the hypothesized parameter value m = 1 is not captured by the CI, one must
(1) Reject Ho: m = 1 at the a = 0.05 % level of significance.
(D) What is xbar?
(1) xbar = [lower CI limit + upper CI limit] / 2
= [ 1.5 + 1.1 ] / 2 = 2.6 / 2 = 1.3
(E) What is the maximum error (E)?
(1) E =
[upper CI limit - lower CI limit] / 2 = Z0.025[s
/ Ö n]
(click me)
= [ 1.5 - 1.1 ] / 2 = 0.4 / 2 = 0.2
(F) What is the standard error (s / Ö n)?
(1) E = Z0.025[s / Ö n] = 1.96 [s / Ö n] = 0.2
(2) [s / Ö n] = 0.2 / 1.96 = 0.102 = SEMEAN
What would be the conclusion in this problem if Ho:
µ = 1.75? FRT(Support)
Ho Reject
Ho (click one)
Example 3
Use Excel to solve this problem. Complete the hypothesis test for the mean of a normally distributed population give the following information:
(a) Ho:
µ £ 1.3; Ha:
µ > 1.3
(click me)
(b) Given: Data: 1.6, 2.6, 0.6, 2.7, 0.5, 2.9, 0.3, 2.9, 0.3, 3.14, 0.06, 3.16, 0.04, 1.6
n = 14
xbar = 1.6
s² = 1.6
a = 0.10
(c) sxbar = s / Ö n
= [ Ö 1.6 ] / [Ö 14 ]
= 1.265 / 3.748 = 0.338
(d) Z or t?
Source of Standard Deviation
|
|
|
Population s |
Sample s |
|
Sample Size |
Small df £ 30 |
Z |
t |
|
|
Large df > 30 |
Z |
Z or t |
(e) If
a = 0.10; t0.10,13
= 1.350 t table
(click me)
(f) t* = [xbar - m ] / [ sxbar]
= [ 1.6 - 1.3 ] / [ 0.338 ]
= 0.89
(g) t* < t a , df
0.89 < 1.350;
FTR ( support )Ho
(h) Small-Sample One-Tailed (right) Tests on the Population Mean
Ho: m £ m o
Ha: m > m o
Reject H o if t* > t a ,df
FTR(Support) H o if t* £ t a ,df
(i) Since at a = 0.10 and df = 13 the sample xbar = 1.6 is statistically close enough to the hypothesized parameter value of 1.3, one can believe that Ho: m£ 1.3 is true; therefore,
(j) FTR (support )Ho (t* < t0.10,14 i.e. 0.89 < 1.35). There is no statistically significant evidence that the mean in this normally distributed population is greater than 1.3.
What would be the conclusion in this problem if Ho:
µ ³ 1.3? FTR(Support)
Ho Reject
Ho (click one)
Example 4
A magazine article claims that the average life time of batteries when used in ordinary toys is not the same as 10 years ago; it claims the average life time of batteries (m) is now 250 minutes. Your firm sells battery powered toys, so the value of this population mean is of vital interest to you. To investigate the article's claim, you randomly select 36 batteries for testing. Your results are:
Use the p-value to draw a conclusion about the null hypothesis. Use Excel to solve this problem.
(1) Ho: m = 250; Ha: µ ¹ 250
(2) ) Given Data: 240, 250, 230, 245, 235, 241, 239, 242, 238, 243, 237, 244, 236, 246, 234, 248, 232, 249, 231, 250, 230, 230, 250, 210, 270, 211, 269, 200, 280, 199, 281, 195, 285, 190, 290,240
xbar = 240
s = 24
a = 0.05
n = 36
sxbar = 4.0
(3) Z* > Z a
|-2.50| > 1.645; Reject Ho
(4) Since at a = 0.05 the sample xbar = 240 is statistically so much smaller than the hypothesized parameter of 250, one can not believe that Ho: µ = 250 is true; therefore, Reject Ho. (|Z*| > Z0.025 i.e. 2.50 > 1.645)
(5) p/2-value = P[Z < - 2.50 ] = 0.5 - P[ - 2.50 < Z < 0 ]
= 0.5 - 0.4938 = 0.0062 Z table
(6) p/2-value < a/2 = 0.0062 < 0.025
p-value < a; 0.0124 < 0.05; Reject Ho
(7) Since the p-value = 0.0124 is less than a = 0.05 , then |Z*| > Z0.05 ( Z* falls in the tail ) and one can not believe that Ho: m = 250 is true; therefore, reject Ho.
Could Ho be rejected in this problem if the value of alpha were unknown? Yes
No
(click one)
Example 5
It is believed that the mean pre-admission test score for students entering the College of Business Administration at NAU is greater than 150. Assume that the scores are normally distributed. A random sample of 27 students yielded a mean score of 157 with a standard deviation of 11.3. Do these data support the belief? Use the p-value. Use Excel to solve this problem.
(a) Data: 157, 155, 159, 161, 153, 163, 151, 164, 150, 165, 149, 167, 147, 168, 146, 169, 145, 170, 144, 171, 143, 172, 142, 173, 141, 174, 140
n = 27
xbar = 157
s = 11.3
(b) sxbar = s / Ö n
= [11.3] / Ö [ 27 ]
= 11.3 / 5.2 = 2.21
(c) Ho:
µ £ 150; Ha:
µ > 150
(click me)
(d) Z or t?
Source of Standard Deviation
|
|
|
Population s |
Sample s |
|
Sample Size |
Small df £ 30 |
Z |
t |
|
|
Large df > 30 |
Z |
Z or t |
(e) t* = [xbar - m ] / [ sxbar]
= [ 157 - 150 ] / [ 2.21 ]
= 3.16
(f) No a value given, thus df = n - 1 = 27 - 1 = 26
(g) Finding p-value using t
(1) go to t table
(2) find df row in problem: df = n - 1 = 27- 1 = 26
(3) locate the value of t* on that row
(4) Note: probabilities are given by the subscript in t a at top of columns
(h) p-value < 0.005; Reject Ho
(i) Rule of Thumb:
(1) Reject Ho if p-value is small (p-value < 0.01 )
(2) In this problem, p-value < 0.005 < 0.01, reject Ho
(j) Most levels of significance ( a ) chosen are > 0.01; therefore, when p-value < 0.01; reject Ho, since p-value would be < a.
(k) Since using the p-value < 0.005 < 0.01 at df = n - 1 = 27 - 1 = 26, the sample xbar = 157 is statistically so much larger than the hypothesized parameter value of 150, one can not believe that Ho: µ £ 150 is true; therefore,
(l) Reject Ho. There is statistically significant evidence that the mean aptitude score is greater than 150.
Excel would give a p-value more exact than that found by looking in a t table.
True
False
(click one)
Go on to Excel
and Equations
or
Go back to Hypothesis
Testing: Activities and Assignments
Please reference "BA501 (your last name) Assignment name and number" in the subject line of either below.
E-mail Dr. James V. Pinto at
BA501@mail.cba.nau.edu
or call (928) 523-7356. Use WebMail for attachments.
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Arizona University
ALL RIGHTS RESERVED