G Listed below are the number of cricket chirps in 1 min and the corresponding temperatures in degrees Fahrenheit. Is there sufficient evidence to conclude that there is a relationship between the number of cricket chirps in 1 min and the temperature? Use a significance level of alpha equals 0.05. Chirps in 1 min 1000 836 847 828 865 968 1120 941 Temperature (degrees Upper F )89.4 73.8 86.7 88.1 81.8 87.3 94.6 90.3 Determine the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0: rho Subscript snot equals0 Upper H 1: rho Subscript sequals0 B. Upper H 0: r Subscript snot equals0 Upper H 1: r Subscript sequals0 C. Upper H 0: r Subscript sequals0 Upper H 1: r Subscript snot equals0 D. Upper H 0: rho Subscript sequals0 Upper H 1: rho Subscript snot equals0 Your answer is correct. Find the value of the correlation coefficient.

what are the statements

step-by-step explanation:

a - base

b - exponent

so,

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.