Question

Maganement at a radio station wants to know the number of songs it plays per hour. A random sample over the past month yields the following counts for songs per hour: 9,13,12,8,7,10,11,10,8,12,13,9,10,12, and 11.A. The radio station wants to claim that they play more than 10 songs per hour. What are the null and alternative hypotheses that test this statement?B. What test willyou use to evalute the alternative hypothesis? What are the conditions of this test? Are they met in this situation?C. Calculate your test statistic and P-value. Show your work.D. What's your critical t value and your conclusion (at a=.05)

Answer 1

we conclude that the radio station does not play more than 10 songs per hour.

Explanation:

We are given the following data set:

9,13,12,8,7,10,11,10,8,12,13,9,10,12,11

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(155)/(15) = 10.34`

Sum of squares of differences = 49.34

`S.D = \sqrt{(49.33)/(14)} = 1.88`

We are given the following in the question:

Population mean, μ = 10

Sample mean,
`\bar{x}` = 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

`H_(0): \mu \leq 10\\H_A: \mu > 10`

We use one-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(10.33 - 10)/((1.88)/(√(15)) ) = 0.6798`

Now,
`t_(critical) \text{ at 0.05 level of significance, 14 degree of freedom } = 1.7613`

Since,

`t_(stat) < t_(critical)`

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.