If n=180 and ˆpp^ (p-hat) = 0.5, construct a 99% Confidence Interval. In order to use technology, you need to compute the number of successes: x = n*ˆpp^

CNNBC recently reported that the mean annual cost of auto insurance is 1031 dollars. Assume the standard deviation is 204 dollars. You take a simple random sample of 63 auto insurance policies.Find the probability that a single randomly selected value is less than 971 dollars.P(X < 971) = Find the probability that a sample of size n=63n=63 is randomly selected with a mean less than 971 dollars.P(M < 971) = Enter your answers as numbers accurate to 4 decimal places. Submit QuestionQuestion 13

A doctor believes women have been getting taller. She knows that 10 years ago the average height of women in her city was 63 inches. The standard deviation is unknown. She randomly samples 8 women from the town and finds their average height to be 64.5 inches, the standard deviation was 6.5. state the hypotheses(H0 and H1) What is your conclusion based on α=0.052tail

CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars. You take a simple random sample of 96 auto insurance policies.Find the probability that a single randomly selected value is less than 969 dollars.P(x<969)=P(x<969)= Find the probability that a sample of size n=96n=96 is randomly selected with a mean less than 969 dollars.P(¯x<969)=P(x¯<969)= Enter your answers as numbers accurate to 4 decimal places.

CNNBC recently reported that the mean annual cost of auto insurance is 1002 dollars. Assume the standard deviation is 291 dollars. You take a simple random sample of 72 auto insurance policies.Find the probability that a single randomly selected value is less than 988 dollars.P(X < 988) = Find the probability that a sample of size n=72n=72 is randomly selected with a mean less than 988 dollars.P(M < 988) = Enter your answers as numbers accurate to 4 decimal places.

CNNBC recently reported that the mean annual cost of auto insurance is 973 dollars. Assume the standard deviation is 234 dollars. You take a simple random sample of 61 auto insurance policies. Find the probability that a single randomly selected value is less than 991 dollars.P(X < 991) =  Find the probability that a sample of size n=61n=61 is randomly selected with a mean less than 991 dollars.P(M < 991) =

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1491 and a standard deviation of 299. The local college includes a minimum score of 1940 in its admission requirements. What percentage of students from this school earn scores that fail to satisfy the admission requirement? P(X < 1940) = Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

CNNBC recently reported that the mean annual cost of auto insurance is 966 dollars. Assume the standard deviation is 294 dollars. You take a simple random sample of 76 auto insurance policies.Find the probability that a single randomly selected value is less than 968 dollars.P(X < 968) = Find the probability that a sample of size n=76n=76 is randomly selected with a mean less than 968 dollars.P(M < 968) = Enter your answers as numbers accurate to 4 decimal places.

A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.07 years, with sample standard deviation s = 0.88 years. However, it is thought that the overall population mean age of coyotes is ? = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use ? = 0.01. (a) What is the level of significance?State the null and alternate hypotheses. H0: ? = 1.75 yr; H1: ? > 1.75 yrH0: ? > 1.75 yr; H1: ? = 1.75 yr    H0: ? = 1.75 yr; H1: ? ≠ 1.75 yrH0: ? < 1.75 yr; H1: ? = 1.75 yrH0: ? = 1.75 yr; H1: ? < 1.75 yr (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The Student's t, since the sample size is large and ? is known.The standard normal, since the sample size is large and ? is unknown.    The standard normal, since the sample size is large and ? is known.The Student's t, since the sample size is…

A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 25 bacteria reveals a sample mean of = 70 hours with a standard deviation of s = 7 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 99% level of confidence. What sample size should you gather to achieve a 0.5 hour margin of error? He would need to sample bacteria.