I don’t understand on what your asking
Explanation:
We are given:
sample size= n = 45
sample mean = x = 61,300
sample standard deviation =18,246
Since the population standard deviation is not known we will use t distribution to find the confidence interval.
Confidence interval = 99%
Degrees of freedom= n - 1 = 45 – 1 = 44
Critical t value = 2.692
The 99% confidence interval will be:![(x-t_{critical}* \frac{s}{ \sqrt{n} }, x+t_{critical}* \frac{s}{ \sqrt{n} }) \\ \\ (61300-2.692* \frac{18246}{ \sqrt{45} }, 61300+2.692* \frac{18246}{ \sqrt{45} }) \\ \\ (53978,68622)](https://tex.z-dn.net/?f=%28x-t_%7Bcritical%7D%2A%20%5Cfrac%7Bs%7D%7B%20%5Csqrt%7Bn%7D%20%7D%2C%20x%2Bt_%7Bcritical%7D%2A%20%5Cfrac%7Bs%7D%7B%20%5Csqrt%7Bn%7D%20%7D%29%20%5C%5C%20%5C%5C%20%2861300-2.692%2A%20%5Cfrac%7B18246%7D%7B%20%5Csqrt%7B45%7D%20%7D%2C%2061300%2B2.692%2A%20%5Cfrac%7B18246%7D%7B%20%5Csqrt%7B45%7D%20%7D%29%20%5C%5C%20%5C%5C%20%2853978%2C68622%29)
Thus, the 99% confidence interval for the population mean is:
53978 to 68622
I unfortunately don’t have anything to answer this question with.
50 dollars at max and Yelp