Consider the following random sample of size eight from a no

Consider the following random sample of size eight from a normal population. Based on the sample, perform an hypothesis test to test the claim that the mean of the population is greater than 100 at ? = 0.05                     

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It has been reported that the average credit card debt foe college seniors is $3262. The student senate at a large university feels that their seniors have a debt much less than this, so it conducts a survey of 50 randomly selected seniors and finds that the average debt is $2995 with a sample standard deviation of $1100. With  ? = 0.05, is the student senate correct?

Solution

sample size= n=8

sample mean =105.625

sample standard deviation =21.61968

Let mu be the population mean

The test hypothesis:

Ho: mu=100 (i.e. null hypothesis)

Ha: mu>100 (i.e. alternative hypothesis)

The test statistic is

t=(xbar-mu)/(s/vn)

=(105.625-100)/(21.61968/sqrt(8))

=0.74

It is a right-tailed test.

The degree of freedom =n-1=8-1=7

The p-value= P(t with df=7 >0.74) =0.2417 (from student t table)

Since the p-value is larger than 0.05, we do not reject the null hypothesis.

So we can not conclude that the mean of the population is greater than 100

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Ho: mu=3262

Ha: mu<3262

The test statistic is

Z=(xbar-mu)/(s/vn)

=(2995-3262)/(1100/sqrt(50))

=-1.72

It is a left-tailed test.

The p-value= P(Z<-1.72) =0.0427 (from standard normal table)

Since the p-value is less than 0.05, we reject the null hypothesis.

So we can conclude that the student senate is correct