Question

3) A random sample of 1001 Americans aged 15 or older revealed that the amount of time spent eating or drinking per day is 1.22 hours, with a standard deviation of 0.65 hours.

a) Suppose a histogram of time spent eating and drinking were right-skewed. Use this result to explain why a large sample size is needed in order to construct a confidence interval for the mean time spent eating and drinking each day. (2 points)

b) There are over 215 million Americans aged 15 and older. Explain why this fact, and the fact that our sample was a true random sample, satisfies the requirements for constructing a confidence interval. (2 points)

c) Construct 90%, 95% and 99% confidence intervals for the mean amount of time Americans aged 15 and older spend eating and drinking per day. Be sure to show all work (or, if using StatCrunch, paste your output to this document) and provide a proper interpretation for each interval

Answer 1

Answer:

Given,

xbar = 1.22

standard deviation = 0.65

a)

Here if the sample is large i.e., > 30, then the sampling distribution of sample mean is approximately normal.

To construct the confidence interval for mean we need sampling distribution of sample mean is approximately normal.so the distribution is strongly skewed & then we need CLT, so the n should be > 30.

b)

Here we need the sample which is < 5% of population is then satisfied.

c)

sample n = 1001

degree of freedom = n - 1

= 1001-1

= 1000

For 90% CI, t(alpha/2,df) = t(0.05 , 1000) = 1.65

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 1.22 +/- 1.65*0.65/sqrt(1001)

= (1.1861 , 1.2539)

For 95% CI, t(alpha/2,df) = t(0.025 , 1000) = 1.96

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 1.22 +/- 1.96*0.65/sqrt(1001)

= (1.1797 , 1.2603)

For 99% CI, t(alpha/2,df) = t(0.005 , 1000) = 2.58

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 1.22 +/- 2.58*0.65/sqrt(1001)

= (1.1670 , 1.2730)