Students are asked about the number of songs they downloaded from a pay for song
ID: 3225964 • Letter: S
Question
Students are asked about the number of songs they downloaded from a pay for songs Website last month. From a random sample of 39 students , the sample mean was 4.7 with a standard deviation of 3.2
Obtain a 95 % confidence interval for the mean number of songs downloaded by the population of all students
Does the population mean lie in the interval obtained in part a ?
According to the above confidence interval is 5 a reasonable value for the population mean number of songs downloaded ?
In a long series of repeated experiments with new samples of 39 students, what percentage of the resulting confidence intervals will contain the true population mean ?
Is there sufficient evidence to say that the population mean number of songs downloaded is less than 6? Use a 5 % significance level
State the null and alternative hypothesis
Find the value of the test statistic
Find the rejection region
State your conclusion
What is the potential type of error based on your conclusion ?
What would constitute type I and type II errors in this example ?
Explanation / Answer
Degree of freedom =n-1=39-1=38
Two tailed Critical t(0.05,38)=2.024
95% confidence interval=xbar-2.024*s/sqrt(n) , xbar+2.024*s/sqrt(n)
=4.7-2.024*3.2/sqrt(39) 4.7+2.024*3.2/sqrt(39)
=(3.66 5.74)
We are 95% confident that this interval contains true population mean
Yes, as 5 is contained by the confidence intevral
95% of the resulting confidence intervals will contain the true population mean
We have to test the null hypothesis H0: µ=6 against alternative Ha:µ<6
As we do not know the population standard deviation , so will use t-score
Test statistic t=(xbar-6)/(s/sqrt(n))=(4.7-6)/(3.2/sqrt(39)) =-2.54
Degree of freedom =n-1=39-1=38
Left tailed Critical t at 0.05 level t(0.05,38) =-1.686
As calculate t<-1.686, we have enough evidence to reject the null hypothesis and conclude that the population mean number of songs downloaded is less than 6