In: Statistics and Probability
.How does cellphone service compare between different cities? The data stored in CellService represents the rating of Verizon and AT&T in 22 different cities (Data extracted from “Best Phones and Services,” Consumer Reports, January 2012, p. 28, 37). Is there evidence of a difference in the mean cellphone service rating between Verizon and AT&T? (Use a 0.05 level of significance)
I WANT THE ANSWER IN EXCEL USING THE DATA ANALYSIS....DO NOT PROVIDE A HAND WRITTEN ANSWER...I'm trying to understand the functions in excel
| City | Verizon | AT&T | 
| Atlanta | 74 | 56 | 
| Austin | 72 | 61 | 
| Boston | 69 | 60 | 
| Chicago | 73 | 55 | 
| Dallas-Fort Worth | 76 | 65 | 
| Denver | 72 | 56 | 
| Detroit | 73 | 63 | 
| Houston | 77 | 59 | 
| Kansas City | 74 | 66 | 
| Los Angeles | 73 | 56 | 
| Miami | 74 | 61 | 
| Milwaukee | 74 | 60 | 
| Minneapolis-St.Paul | 71 | 60 | 
| New York | 71 | 57 | 
| Philadelphia | 71 | 63 | 
| Phoenix | 76 | 62 | 
| San Diego | 74 | 60 | 
| San Francisco | 73 | 53 | 
| Seattle | 72 | 59 | 
| St. Louis | 73 | 64 | 
| Tampa | 73 | 67 | 
| Washington D. C. | 71 | 60 | 
Step(1)-
State the null hypothesis
Step(2)-
put the data in excel and go to data analysis option and apply the t-test for two different mean at 0.05 level of significance.
here are given the screenshot of working.
Step(3)-
the following output got

step(3)-
Result-
| Verizon | AT &T | |
| Mean | 73.0 | 60.1364 | 
| Variance | 3.5 | 13.4567 | 
| Observations | 22.0 | 22.0000 | 
| Hypothesized Mean Difference | 0.0 | |
| df | 31.0 | |
| t Stat | 14.6 | |
| P(T<=t) one-tail | 0.0 | |
| t Critical one-tail | 1.7 | |
| P(T<=t) two-tail | 0.0 | |
| t Critical two-tail | 2.0 | 
here we see that p-value is 0.00<0.05 its mean there is strong evidence against the null hypothesis hence we reject the null hypothesis therefore there is a significant difference b/w means of 22 cities for verizon and AT&T.
Note- if it helps you then please appreciate the work ,thumps up.