Stat 217 – HW 5 Solutions

2) Generation M – 10 pts

The generation of children raised in the 1990s and 2000s has been dubbed “Generation M” because of the impact of media on their lives. The Kaiser Family Foundation conducted an extensive study to quantify how much time teenagers spend with various types of media. They gathered data from what they describe as a “nationally representative sample” of 2032 teenagers. One finding was that 31% of the respondents had a computer in their bedroom.

(a) The proportion of all Americans teens that have a computer in their room.

(b) Since n = 2032, if we example n × = 2032(.31) = 629.9 > 10 and n × (1-) = 1402.1 > 10, we see the sample size condition is equally met.

We are also told this was a “nationally representative sample” so we will take their word on that.

(c) Using the Test of Significance Calculator applet we have

“By hand”: + 2 .31 + 2Ö(.31(.69)/2032) = .31 + 2(.0102) = .31 + .02 = (.29, .33)

We are 95% confident that between 29% and 33% of all American teens have a television in their room.

(d) margin of error = (.33-.29)/2 = .02

(e) The new interval from the applet is:

Half-width = (.338-.282)/2 = .028

Or “by hand” margin of error » 2 = .029

The margin of error is larger with the smaller sample size.

(f) with n = 2032: 1/Ö2032 = .022

With n = 1016: 1/Ö1016 = .031

These are pretty similar to the values we found above.

3) Activity 15-10 (p. 304-5) – 12 pts

a.

Approximately 38% of candy bars weight between 2.18 oz and 2.22 oz.

b. Yes – the CLT applies because the population has a normal distribution as long as your sampling method behaves like a simple random sample.

c. The CLT says the sample means will be normally distributed with mean 2.20 ounces and standard deviation =.

d.

Student guess. This value should be greater than the answer to part a.

e.

Approximate 73.6% of samples will have a sample mean between 2.18 and 2.22

This probability is indeed larger than the probability we found in part a.

f. Student guess – they should guess that the probability will increase if the sample size is increased to 40 because this will decrease the standard deviation which will concentrate more area under the curve near the middle (mean).

g. Now the standard deviation of the sample means is .04/= .006325, so the distribution is N(2.2, .006325).

Approximate 99.8% of samples will have a sample mean between 2.18 and 2.22

This is larger that our answer in part e.

h. The calculations in part f remain approximately correct regardless of the distribution of candy bar weights because the sample size was large (40 > 30).

4) Activity 15-17 (p. 307) - 4 pts

a. The CLT says the sampling distribution will be approximately N(15, 4/=.566)

Approximately .67% of samples will have a sample mean above 16.40%.

b. Yes – this provides strong evidence that the mean tip percentage is actually greater than 15%, because if it were 15% or less, the chance that we would find a random sample or 50 tables with an average tip percentage of at least 16.4% is less than 1% - so it is extremely unlikely.

c.

This does not provide strong evidence that the population mean tip percentage is less than 15% because if the population mean percentage is 15% (or more) – we would expect to see a random sample of 50 tables with an average percentage tip of 14.4% or less almost 15% of the time, which is not rare.