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) 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 =
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.
part 1. You measure 33 watermelons' weights, and find they have a mean weight of 55...
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