Find a. the mean; \( \mathbf{b} \). the deviation from the mean for each data item; and \( \mathbf{c} \). the sum of the deviations in part (b) for the following group of data items. \( 61,91,76,61,106 \) a. The mean is (Simplify your answer.)

standard deviation 3 a normal distribution random variable with mean 23 and that the probability \[ P(X>x)=0.66 \]

In a one-sample t-test, which of the following is true about the null hypothesis? (B) The sample mean is greater than the population mean (C) The sample mean is not equal to the population mean (D) The sample size is larger than 30

A Cramer's V of .90 would be indicative of a weak of relationship (A) True (B) False

Define un diagrama de dispersión y describe su propósito en la estadística.

Data from the Office for National Statistics show that the mean age at which men in Great Britain get married was 32.5 . A news reporter noted that this represents a continuation of the trend of waiting until a later age to wed. A new sample of 47 recently wed British men provided their age at the time of marriage. This data is contained below. Age- \( 26,36,27,30,28,26,36,33,25,30,31,34,26,33,29,32,30,39 \), \( 34,27,30,30,35,26,28,29,36,28,28,26,38,25,35,40,26,35,35,27 \), \( 33,29,28,33,38,39,35 \) Do these data indicate that the mean age of British men at the time of marriage exceeds the mean age in 2013 ? Test this hypothesis at a= 0.01. What is your conclusion? Use the obtained rounded values in your calculations. Sample mean: Sample standard deviation: t-value: p-value (Upper Tail):