A random sample of 25 households without health insurance had a mean annual inco

Question

A random sample of 25 households without health insurance had a mean annual income of $37,850 with a standard deviation of $25,554 At the 0.05 level of significance, text the claim that the mean annual income of households without insurance is not $50,000(Based on results of a study by the census Bureau.) What is the null hypothesis? What is the alternative hypothesis? What is the test statistic of confidence interval? (show all justification.) What is your conclusion in terms your parents would understand?

Explanation / Answer

t-test For Single Mean
Set Up Hypothesis
Null, H0: annual income of households without insurance U=50000
Alternate, H1:annual income of households with insurance U!=50000
Test Statistic
Population Mean(U)=50000
Sample X(Mean)=37850
Standard Deviation(S.D)=25554
Number (n)=25
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =37850-50000/(25554/Sqrt(25))
to =-2.377
| to | =2.377
Critical Value
The Value of |t | with n-1 = 24 d.f is 2.064
We got |to| =2.377 & | t | =2.064
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != -2.3773 ) = 0.0258
Hence Value of P0.05 > 0.0258,Here we Reject Ho

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=37850
Standard deviation( sd )=25554
Sample Size(n)=25
Confidence Interval = [ 37850 ± t a/2 ( 25554/ Sqrt ( 25) ) ]
= [ 37850 - 2.064 * (5110.8) , 37850 + 2.064 * (5110.8) ]
= [ 27301.309,48398.691 ]

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