Twelve residents from the city of Rocklin were randomly selected and asked "How many TVs are in your household?". The following data were obtained: 2, 3, 3, 1, 2, 5, 3, 4, 1, 2, 4, and 3. According to Nielsen Media Research the national average is 2.7 TVs per household. Is this sufficient sample evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average of 2.7 TVs per household? Use ! = 5% and assume that the number of TVs in Rocklin households is normally distributed.

There isn't enough evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average .

Step-by-step explanation:

This is a one sample mean test ;

H0 : μ = 2.7

H1 : μ > 2.7

Given the data :

2, 3, 3, 1, 2, 5, 3, 4, 1, 2, 4, 3

Sample size, n = 12

The sample mean, xbar = ΣX / n = 33/12 = 2.75

The sample standard deviation, s = 1.215 ( from calculator)

The test statistic :

(xbar - μ) ÷ (s/√(n))

T = (2.75 - 2.7) ÷ (1.215/√(12))

T = 0.05 / 0.3507402

T = 0.1426

The critical value from Tscore ;

df = 12 - 1 = 11

Critical value = 1.796

Since ; Test statistic < Critical value ;WE fail to reject the Null and conclude that there isn't enough evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )

just sum

10 + 10 = 20