In a survey conducted by a marketing agency, 231 out of 1000 adults 19 years of age or older confessed to bringing and using their cell phone every trip to the bathroom (confessions included texting and answering phone calls).
a. What is the point estimate?
b. Construct a 95% confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom.
c. Interpret your confidence interval.
Determine the t-value in each of the following.
a. Find the t-value such that the area in the right tail is 0.2 with 16 degrees of freedom.
b. Find the critical t-value that corresponds to 60% confidence with 25 degrees of freedom.
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 939 people age 15 or older, the mean amount spent eating or drinker per day is 1.9 hours with a standard deviation of 0.55 hour.
a. Find a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.
b. Interpret your interval.
Please show all work for the following problems. If a calculator is used, include the calculator function used as well as the values entered. (Example: normalcdf
Answer:
1.
Given,
x = 231
n = 1000
a)
Point estimate = p^ = x/n
= 231/1000
= 0.231
b)
Here for 95% CI, z value is 1.96
CI = p^ +/- z*sqrt(p^(1-p^)/n)
substitute values
= 0.231 +/- 1.96*sqrt(0.231(1-0.231)/1000)
= 0.231 +/- 0.026
= (0.231 - 0.026 , 0.231 + 0.026)
= (0.205 , 0.257)
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In a survey conducted by a marketing agency, 231 out of 1000 adults 19 years of...