Question

In: Statistics and Probability

Random samples of people ages 15−24 and of people ages 25−34 were asked about their preferred...

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 Email 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.     

Solutions

Expert Solution

using excel>Addin>phstat>multiple sample test

we have

Chi-Square Test
Observed Frequencies
Preferred Communication Method Calculations
age Cell phone Instant message Email 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 Email 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


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