Calculating P(U-3 < 4.33 sqrtV) Using t Distribution

In summary, the problem asks to find the probability of U being less than 4.33 times the square root of V, given that U and V are independent and have specific distributions. The solution involves understanding the properties of the normal and chi-square distributions, and using them to find the appropriate probabilities. The use of R to calculate the probability is also suggested.
  • #1
dim&dimmer
22
0
t distribution??

Homework Statement


If U and V are independent, U being distributed N(3,16) and V being distributed as chi-square on 9 degrees of freedom, find

P(U-3 < 4.33 sqrtV).


Homework Equations





The Attempt at a Solution


Well I don't even know where to start! Does U-3 suggest the mean is 0? Is the whole distribution of U shifted left by 3? Does that give a t distribution?
I am not asking for the answer but some direction would be appreciated, I just can't seem to understand this at all.
 
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  • #2


U-3 is distributed N(0,16).
 
  • #3


Ok, I understand this and also can reason that the probability would be very high, just by thinking of the graphs of the distributions, however I am still at a loss as to how to compute the probability. I use R to do the calculations, is it some kind of t test or do I put in the distributions and their parameters and just use R to calculate the probability, if so... is the R code just the same as in the question, that is, P(U-3 < 4.33 sqrtV)
 

1. What is the formula for calculating P(U-3 < 4.33 sqrtV) using t distribution?

The formula for calculating P(U-3 < 4.33 sqrtV) using t distribution is:P(U-3 < 4.33 sqrtV) = t(4.33) * sqrt(n-1) * sqrt(V) / (n-1 + t^2(4.33)), where t(4.33) is the t-value for a given confidence level and n is the sample size.

2. How do you interpret the calculated value for P(U-3 < 4.33 sqrtV)?

The calculated value for P(U-3 < 4.33 sqrtV) represents the probability that the true population mean U is within 4.33 standard deviations of the sample mean, given a certain level of confidence.

3. What is the significance of using t distribution in this calculation?

T distribution is used in this calculation because the sample size (n) is small (less than 30) and the population standard deviation (sigma) is unknown. T distribution takes into account the added uncertainty due to the small sample size and provides a more accurate estimate of the true population mean.

4. How can I determine the appropriate t-value to use in the calculation?

The appropriate t-value to use in the calculation depends on the desired confidence level and the degrees of freedom (n-1). A table of t-values or a statistical software can be used to find the corresponding t-value for a given confidence level and degrees of freedom.

5. Can this formula be used for any type of data?

This formula can be used for data that is normally distributed and when the sample size is less than 30. If the data is not normally distributed or the sample size is larger than 30, alternative methods of calculating confidence intervals should be used.

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