The General Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 2.98 with a standard deviation of 2.66. Assume that in a sample of 40 teenagers, the sample standard deviation of daily TV time is 1.9 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

Yes, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

Step-by-step explanation:

H0 : σ² = 2.66²

H1 : σ² < 2.66²

X²c = (n - 1)*s² ÷ σ²

sample size, n = 40

Sample standard deviation, s = 1.9

X²c = ((40 - 1) * 1.9²) ÷ 2.66²

X²c = 140.79 ÷ 7.0756

X²c = 19.897

Using a confidence level of 95%

Degree of freedom, df = n - 1 ; df = 40 - 1 = 39

The critical value using the chi distribution table is 25.6954

Comparing the test statistic with the critical value :

19.897 < 25.6954

Test statistic < Critical value ; Reject the Null

Hence, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130