Question

A classic rock radio station claims to play an average of 50 minutes of music every hour. However, it seems that every time you turn to this station, there is a commercial playing. To investigate their claim, you randomly select 36 different hours during the next month and record what the radio station plays in each of the 36 hours. Here are the number of minutes of music in each of these hours:
44 49 45 51 49 53 49 44 47 50 46 48
(a) Construct and interpret a 90% confidence interval for the average number of minutes of music in each hour.
(b) What can you say about the radio station's claim?

Answer 1

Answer:

a) $47.917-1.796\frac{2.811}{\sqrt{12}}=46.460$    

$47.917+1.796\frac{2.811}{\sqrt{12}}=49.374$    

b) Null hypothesis: $\mu =50$

Alternative hypothesis: $\mu \neq 50$

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We have the following data: 44 49 45 51 49 53 49 44 47 50 46 48

And we can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And replacing we got:

$\bar X= 47.917, s= 2.811$

$\mu$ population mean (variable of interest)  

n=12 represent the sample size  

Part a: Calculate the confidence interval

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)  

The degrees of freedom are given:

$df= n-1= 12-1=11$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,11)".And we see that $t_{\alpha/2}=1.796$

Now we have everything in order to replace into formula (1):

$47.917-1.796\frac{2.811}{\sqrt{12}}=46.460$    

$47.917+1.796\frac{2.811}{\sqrt{12}}=49.374$    

Part b

For this case the system of hypothesis are:

Null hypothesis: $\mu =50$

Alternative hypothesis: $\mu \neq 50$

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.