Respuesta :
Answer:
We reject the null hypothesis and fail to accept it and update that estimate that typical teenager sent does not 67 text messages per day.
Step-by-step explanation:
We are given the sample:
51, 175, 47, 49, 44, 54, 145, 203, 21, 59, 42, 100
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{990}{12} = 82.5[/tex]
Sum of squares of differences = 992.25 + 8556.25 + 1260.25 + 1122.25 + 1482.25 + 812.25 + 3906.25 + 14520.25 + 3782.25 + 552.25 + 1640.25 + 306.25 = 3539.363636
[tex]S.D = \sqrt{\frac{3539.363636}{11}} = 59.5[/tex]
We are given the following in the question:
Population mean, μ =67
Sample mean, [tex]\bar{x}[/tex] = 82.5
Sample size, n = 12
Alpha, α = 0.05
Sample standard deviation, s = 59.5
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 67\\H_A: \mu > 67[/tex]
We use One-tailed t test to perform this hypothesis.
b) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{82.5 - 67}{\frac{59.5}{\sqrt{11}} } = 0.864[/tex] Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 11 degree of freedom } = 1.795[/tex]
a) Since,
[tex]t_{stat} < t_{critical}[/tex]
We reject the null hypothesis and fail to accept it and update that estimate that typical teenager send more than 67 text messages per day.
