Question

10.21 How do the ratings of TV and Internet services compare? The file Telecom contains the rating of 14 different providers. (Data extracted from "Ratings: TV, Phone, and Internet Services," Consumer Reports, May 2014. pp. 28-29.) the mean service rating between TV and Internet services? a. At the 0.05 level of significance, is there evidence of a difference b. What assumption necessary about the population distribution in order to perform this test? a Use a graphical method to evaluate the validity of the assumption in (a). d. Construct and interpret a 95 % confidence interval estimate of the difference in the mean service rating between TV and Internet services.

Provider	TV	Phone
Verizon FIOS	74	78
WOW	74	83
Bright House Networks	70	77
A T & T U-verse	68	72
Cox	67	74
SuddenLink	66	81
Cablevision/Optimum	64	75
Insight	64	73
RCN	62	71
Comcast	61	70
TimeWarner	60	71
Charter	59	71
Mediacom	54	63

Answer 1

A.

n₁ = 13

$\bar{x}$₁ = 64.846

s₁ = 5.843

n₂ = 13

$\bar{x}$₂ = 73.769

s₂ = 5.199

Let's conduct a hypothesis for equality of variances of the two population:

H₀: $\sigma$₁² = $\sigma$₂²

H₁: $\sigma$₁²  $\neq$  $\sigma$₂²

Formula Used:

F test for equality of Two Variances F = 2 variance of sample l where S1 2 variance of sample 2 S2 size of the sample taken from population size of the sample taken from population 2 1 degree of freedom from sample 1 1 degree of freedom from sample 2 n2 30

F = 1.263

p-value for F_12,12 = 0.346 > 0.05 i.e H₀ can't be rejected and hence we can say that variances of the two populations' are equal.

Now,

Let's test the hypothesis that for the difference in mean service ratings between TV and internet service providers.

H₀: $\mu$₁ =  $\mu$₂

H₁: $\mu$₁$\neq$  $\mu$₂

Formula Used:

Populations are normally distributed The population variances are assumed equal (o2= o22= o), the two sample variances and pool them to estimate a2. the test statistic is a t value with (n, + n, -2) degrees of freedom so use (,-X)-(4-4) t: 1 1 2 The pooled estimator of o', denoted by s, is defined by (X-X)-(X,-X,) (-1)s + (n, -1) 2i i-1 2 i-1 (-1)+(n,-1emr n-2 7 11 ΧυριΤKGP Il

Assumptions: Populations are normally distributed and the populations' variances are equal (proved above)

t = −4.114

df = 13 + 13 - 2

= 24

p-value = 0.0004 < 0.05 i.e. We can reject H₀ and hence we can say that there is a significant difference between mean service ratings of TV and internet service providers.

B.

Assumptions: Populations are normally distributed

C.

To be assured of the assumption made for the test in part A, we can draw the normal Q-Q plot of the data to see whether the data is normally distributed or not.

Normal QQ Plot for TV service providers:

Normal Q-Q Plot C O C IC IC -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles 99 09 0/ Sample Quantiles

Normal QQ Plot for Internet service providers:

Normal Q-Q Plot -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles O C C 0/ S9 08 G/ Quantiles Sample

As we can see most of the points lie close to the line, thus we can say that samples are normally distributed and hence if samples of such small sizes are normally distributed the population from which they are sampled will also be normally distributed.

D.

Formula:

A 100(1 a)% confidence interval for u 1 (X-X2)tta/2.nm,-2 2 is: 1 Sm n2 Where t has (n, + n2-2) degrees of freedom, and |(4-1)s+(n-1) t 2-2 2 2

Using the above formula:

CI = [-13.4, −4.446]

Interpretation: CI implies that if a large no. of samples are withdrawn for TV and internet rating and the confidence interval is constructed for all of these, the population mean difference between the ratings of TV and internet service providers will lie in 95% of those CI.

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