Respuesta :
Answer:
we conclude that the radio station does not play more than 10 songs per hour.
Step-by-step explanation:
We are given the following data set:
9,13,12,8,7,10,11,10,8,12,13,9,10,12,11
[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{155}{15} = 10.34[/tex]
Sum of squares of differences = 49.34
[tex]S.D = \sqrt{\frac{49.33}{14}} = 1.88[/tex]
We are given the following in the question:
Population mean, μ = 10
Sample mean, [tex]\bar{x}[/tex] = 10.34
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation, s = 1.88
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 10\\H_A: \mu > 10[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{10.33 - 10}{\frac{1.88}{\sqrt{15}} } = 0.6798[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = 1.7613[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the radio station does not play more than 10 songs per hour.
