In: Statistics and Probability
Random samples of people ages 15−24 and of people ages 25−34 were asked about their preferred method of (remote) communication with friends. The respondents were asked to select one of the methods from the following list: cell phone, instant message, e-mail, other.
Preferred Communication Method | |||||
Age | Cell Phone | Instant Message | Other | Row Total | |
15-24 | 47 | 45 | 3 | 5 | 100 |
25-34 | 43 | 28 | 16 | 13 | 100 |
Column Total | 90 | 73 | 19 | 18 | 200 |
(i) Make a cluster bar graph showing the percentages in each age group who selected each method. (In the graphs, cyan represents ages 15-24, and gray represents ages 25-34.)
(ii) Test whether the two populations share the same proportions
of preferences for each type of communication method. Use
α = 0.05.(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Different proportion of communication
preference for each age group.
H1: Different proportion of communication
preference for each age group.H0: Different
proportion of communication preference for each age group.
H1: Same proportion of communication preference
for each age
group. H0: Same
proportion of communication preference for each age group.
H1: Same proportion of communication preference
for each age group.H0: Same proportion of
communication preference for each age group.
H1: Different proportion of communication
preference for each age group.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
binomialStudent's t normalchi-squareuniform
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
p-value > 0.1000.050 < p-value < 0.100 0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that the proportion of age groups within each communication preference is not the same.At the 5% level of significance, there is insufficient evidence to conclude that the proportion of age groups within each communication preference is not the same.
using excel>Addin>phstat>multiple sample test
we have
Chi-Square Test | ||||||||||
Observed Frequencies | ||||||||||
Preferred Communication Method | Calculations | |||||||||
age | Cell phone | Instant message | Other | Total | fo-fe | |||||
15-24 | 47 | 45 | 3 | 5 | 100 | 2 | 8.5 | -6.5 | -4 | |
25-34 | 43 | 28 | 16 | 13 | 100 | -2 | -8.5 | 6.5 | 4 | |
Total | 90 | 73 | 19 | 18 | 200 | |||||
Expected Frequencies | ||||||||||
Preferred Communication Method | ||||||||||
age | Cell phone | Instant message | Other | Total | (fo-fe)^2/fe | |||||
15-24 | 45 | 36.5 | 9.5 | 9 | 100 | 0.088889 | 1.979452 | 4.447368 | 1.777778 | |
25-34 | 45 | 36.5 | 9.5 | 9 | 100 | 0.088889 | 1.979452 | 4.447368 | 1.777778 | |
Total | 90 | 73 | 19 | 18 | 200 | |||||
Data | ||||||||||
Level of Significance | 0.05 | |||||||||
Number of Rows | 2 | |||||||||
Number of Columns | 4 | |||||||||
Degrees of Freedom | 3 | |||||||||
Results | ||||||||||
Critical Value | 7.814728 | |||||||||
Chi-Square Test Statistic | 16.58697 | |||||||||
p-Value | 0.000859 | |||||||||
Reject the null hypothesis |
(ii)
a) the level of significance is 0.05
H0: Same proportion of communication preference for
each age group.
H1: Different proportion of communication preference for
each age group.
(b) the value of the chi-square statistic for the sample. is
16.587
yes ,all the expected frequencies greater than 5
What sampling distribution will you use?
normal
What are the degrees of freedom?
3
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
p value is 0.0001
p-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value ≤ α, we reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that the proportion of age groups within each communication preference is not the same