Stat 217 – HW 4 Solutions

1) Survey- 3 pts

2) Activity 12-14 (p. 250) – 8 pts

(a)

z = (25-22.5)/2.5 = -1 with probability below .1587

z = (27.5-25)/2.5 = 1, with probability below .8413

probability between = .8413 -.1587 = .6826

(b) By the empirical rule, about 95% of all male Sheltie’s heights fall between 15-2(1.5)=12 and 15+2(1.5)=18 inches (within 2 standard deviations).

(c) The top 10% (or bottom 90%) corresponds to z = 1.28 (table or technology), so then solving for height, we find:

1.28 = (height – 15)/1.5 height = 16.92 inches

(d) Sheltie: z =(18-15)/1.5 = 2.00

Shepherd: z = (28-25)/2.5 = 1.20

So the Sheltie height is further from the mean compared to its peers. The Sheltie having a height of 18 inches is more unusual than the German Shepherd having a height of 28 inches.

3) Activity 13-14 (p. 271-2) – 14 pts

(a) A parameter since it applies to “all American households”, p

(b) Does the Central Limit Theorem apply?

With n = 80,000 and p = 1/3, we have n × p = 80,000(1/3) = 26666.67 and n × (1-p) = 80,000(2/3) = 53333.33. Both are clearly above 10. We are told that it was a national sample so the random sampling is not clearly stated but could be presumed.

Since the Central Limit Theorem applies, it tells us that the sampling distribution of the sample proportion will be approximately normal with mean equal to 1/3 and standard deviation equation to Ö (1/3)(2/3)/80000 = .00167.

Here is the sketch:

(c) We expect 95% of all sample proportions to fall within 2 standard deviations of the mean. Two standard deviations = 2 × .001667 = .0033.

Mean – 2 × std dev = .3333 - .0033 = .3300

Mean + 2 × std dev = .3333 + .0033 = .3367

We expect 95% of all sample proportions to fall between .3300 and .3367.

(d) The interval is so narrow because the sample size (n = 80,000) is very large.

(e) This is a statistic since it describes the sample.

(f) z = (.316 - .3333)/.001667 » -10.4

(g) These data provide very strong evidence that the population proportion is not one-third. We would get a sample proportion as small as .316 pretty much never if p = 1/3, so we will reject p = 1/3, and conclude p differs from 1/3.

4) Continuing the previous problem – 10 pts

(h) Does the Central Limit Theorem apply?

With n = 800 and p = 1/3, we have n × p = 800(1/3) = 266.67 and n × (1-p) = 800(2/3) = 533.33. Both are clearly above 10. We are told that it was a national sample so the random sampling is not clearly stated but could be presumed.

Since the Central Limit Theorem applies, it tells us that the sampling distribution of the sample proportion will be approximately normal with mean equal to 1/3 and standard deviation equation to Ö (1/3)(2/3)/800 = .0167.

Here is the sketch:

(f) z = (.316 - .3333)/.0167 » -1.04

(i) Using the Normal Probability Model to determine the probability of obtaining a sample proportion of at most .316 ( < .316) when n = 800.

This sample would not provide convincing evidence that the population proportion was less than 1/3.

(j) Provide a careful (long-run relative frequency) interpretation of the probability you found in (i).

If we were to repeatedly take random samples of 800 households from this population (with p = 1/3), about 15% of those samples would have .316.

(k) If the sample proportion had been more like .25, then we would have rejected /13 as a plausible value for the population proportion.