Question

How common are financial cost or contractual constraints associated with smartphone ownership? A survey of smartphone owners found that 48% of the 18 to 29 year old customers, 38% of the 30 to 49 year old customers, 25% of the 50 to 64 year old customers, and 19% of the 65 year old plus customers have reached the maximum amount of data they are allowed as part of their plan, at least on occasion. Suppose the survey was based on 200 smartphone users in each age category.

1. At the 0.05 level of significance, is there evidence of a difference among the age groups in the proportion of smartphone owners who have reached the maximum amount of data they are allowed to use as part of their plan, at least on occasion?
2. Determine the p-value in (a) and interpret its meaning.
3. If appropriate, use the Maracuilo procedure at .05 to determine which age groups differ. State which ones differ and why statistically.

Answer 1

Answer: ---- Date: ----19/5/2019

a) The findings of a survey of smartphone owners to see whether they are allowed to use as part of their plan are given in a 2x4 contingency table as: Ownership Age group (in years) Total 1節20 30-49 50-64 65 or older Yes 96 76 50 38 124 150 182 104 No 540 200 200 200 Total 800 200 To test the significant difference among the age groups in the proportion of smartphone owners who have reached the maximum amount of data they are allowed to use as part of their plan, Chi Square (z') Test for Differences among more than two proportions will be used. The null and alternate hypotheses are given as: There is no difference among the population proportions. H:Not all, are equal, (j-1,2,.. 4) There is difference among the population proportions.

The procedure for Chi-Square Test for Differences among more than two proportions is given below 1. Open Microsoft Exce 2. Click File-Open to open PHStat.xlam file. 3. A security message will appear on the screen as: Microsoft Excel Security Notice Microsoft Office has identified a potential security concern. Warning: It is not possible to determine that this content came from a trustworthy source. You should leave this content disabled unless the content provides critical functionality and you trust its File Path: C:. ADesktop Evelyn Stats New folder PHStat 4PHStat2.xlam Macros have been disabled. Macros might contain viruses or other security hazards. Do not enable this content unless you trust the source of this file. More information Enable Macos Disable Macros

Click "Enable Macros" to enable the PHStat Add-Ins work properly 4. Click Add-Ins → PHStat and then select Multiple-Sample Test → Chi-Square Test Insert Page Layout Formulas Home Data Review PHStat Data Preparation Descriptive Statistics Decision-Making Probability & Prob. Distributions Sampling Confidence Intervals 65 or over Sample Size 50 38 162 150 One-Sample Tests Two-Sample Tests Multiple-Sample Tests Chi-Square Test... Kruskal-Wallis Rank Test... Control Charts Levene Test... Regression One-Way ANOVA... Utilities About PHStat... Help for PHStat

5. A dialogue box will appear. Enter 0.05 in the Level of Significance, 2 in Number of Rows and 4 in Number of Columns and Tick Marascuilo Procedure and then OK Chi-Square Test 3003 Level of Significance: o.0s 0.05 Number of Rows: 决 Number of Columns: 4 Output Options itte Marascuilo Procedure Help Cancel OK

A new Chi Square sheet will appear. Now enter row, column and other specific fields in the Observed Frequencies table and the Expected Frequencies table is automatically generated as given below Chi-Square Test Observed Frequencies Column variable Total C4 C3 C1 C2 Row variable 260 76 R1 162 150 124 104 R2 800 200 200 200 200 Total Expected Frequencies Column variable Total C3 C1 C2 Row variable 260 R1 135 135 135 135 R2 800 200 200 200 200 Total

Calculations fo-fe 31 15 27 27 15 -31 (fo-fe) 2/fe 14.78462 1.861538 3.461538 11.21538 7.118519 0.896296 1.666667 5.4

Level of Significance 0.05 Number of Rows 2 Number of Columns 4 Degrees of Freedom Results Critical Value 7.814728 Chi-Square Test Statistic 46.40456 p-Value 4.65E-10 Reject the null hypothesis Expected frequency assumption is met.

The calculated value of the Chi-Square Test Statistic and the Critical Value at 5% level of significance with 3 degrees of freedom is given in the above table: χ2TA,-146.403 χ0.os =17,815 Since, calculated value of Chi-Square test statistic is more than the critical value at 5% level of significance with 3 degrees of freedom. Thus, null hypothesis is rejected. In other words, there is significant difference among the age groups in the proportion of smartphone owners who have reached the maximum amount of data they are allowed to use as

b) Compute the p-value as followS: The value of test statistic is 46.405 with 3 degrees of freedom. p-value P(zi 246.405) CHIDIST (46.405,3) 4.65134x10Using Excel function 0.0000 The p-value of the test is approximately zero It means that, the probability of obtaining the value of test statistic greater than or equal to 44.6503 when the null hypothesis is true is 0.0000.

c) Since, the nul hypothesis is rejected in above case, therefore, it is appropriate to check which two proportions differ significantly. This would be done by using the Marascuilo procedure which is already obtained in the Marascuilo sheet in a. as shown below:

Marascuilo Procedure Level of Significance 0.05 Square Root of Critical Value 2.795483483 Sample Proportions Group 1 0.48 Group 2 0.38 Group 3 0.25 0.19 Group 4 MARASCUILO TABLE Proportions Absolute Differences lCritical Ran 0.1 0.137689977 Not significant 0.23 0.130686984 Significant 0.29 0.125563584 Significant Group 1 - Group 2 Group 1- Group 3 Group 1- Group 4 0.13 0.128577046 Significant Group 2-Group 3 0.19 0.123366051 Significant Group 2-Group 4 Group 3-Group 4 0.06 0.115497794 Not significant It is clear from the above table that there is significant difference between the age groups (18-29 and 50-64), (18-29 and 65 or above), (30-49 and 50-64), (30-49 and 65 or above)