Question

part 1.
You measure 33 watermelons' weights, and find they have a mean weight of 55 ounces. Assume the population standard deviation is 9.2 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

part 2.
In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a zz score of 1.96 for your computations.)

a. (.728, .772)
b.(.723, .777)
c.(.718, .782)
d.(.713, .878)
e.(.665, .835)

Answer 1

Answer

(1) We have mean = 55, standard deviation = 9.2 and sample size is 33

we have to calculate 99% confidence interval

z score for 99% confidence interval is 2.58 (using z distribution table)

Formula for the confidence interval is given as

setting the given values, we get

this gives us

CI = (50.87, 59.13) (rounded to two decimals)

(2) we have sample proportion =

750/10000.75

sample size is n = 1000

we have to find 95% confidence interval for the proportion

we have to use z = 1.96

Formula fo rthe confidence interval is given as

setting the given values, we get

we get

CI = (0.723, 0.777)

So, option B is correct.