Problem: According www. textrequest, an American between 18-24 sends and receives about 128 text messages per day. Assume that the population standard deviation is 30 for this age group. To get an idea about the true population mean you take a survey of 100 adults in that age group and find that their average is 135. Create a 90% confidence interval for the population average for this age group.

(130.08, 139.92)

Step-by-step explanation:

Given that:

Mean (m) = 135

Standard deviation (σ) = 30

Sample size (n) = 100

α = 90%

The confidence interval value is obtained using the relation:

Mean ± Zcritical * (standard deviation / sqrt(n))

Zcritical at 90% = 1.64

standard deviation / sqrt(n) = 30 / sqrt(100)

= 30 / 10 = 3

Hence,

135 ± 1.64 * (3)

Lower limit : 135 - (1.64 * 3) = 130.08

Upper limit : 135 + (1.64 * 3) = 139.92

Hence,

Confidence interval = (130.08, 139.92)

step-by-step explanation:

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