Stat 217 - Review Problems Solutions

Activity 12-15

a. z = (13.9-12.5)/1.5 = .93. At 3 months, Benjamin Chance’s weight was about 9/10 of a standard deviation above the average.

b. Pr(weight > 13.9)

We must assume that 3-month old American baby weights are normally distributed.

c. By the empirical rule, to be in the middle 68% of weights, his weight would need to be within one standard deviations of the mean, so within 17.25 ± 2 = 15.25 lbs and 19.25 lbs.

Activity 13-15 (p. 272):

a.

label: sample proportion that are bird owners

The CLT says this sampling distribution will be approximately normal, centered at .05, with a standard deviation of Ö(.05(.95)/80000 = .000771.

b. This standard deviation is much smaller because we are assuming π is smaller (.05 rather than .333, further away from .5, see Activity 13-7).

c. z = (.046-.05) / .000771 = -5.19. This is a very unusual z-score. Pr(Z< -5.19) ≈ 0 – so the survey does provide convincing evidence that the population proportion who own a pet bird is not 5% (we observed a sample result that pretty much never happens when π = .05, so we are convinced π ≠ .05).

Activity 14-11abe (p. 293)

a. No – it would not be surprising to obtain 30.4 mpg for one tankful. This value is well within one standard deviation of the mean (z = (30.4 -31)/3 = -.2)

b. The CLT predicts the sampling distribution of the sample means would be (approximately) normal with mean 31 mpg and standard deviation 3/= .5477 mpg. According to our sketch 30.4 would not be an unusual mean value for a sample of 30 cars to obtain.

c. The CLT predicts a sampling distribution that would be (approximately) normal with mean 31 mpg and standard deviation 3/= .3873 mpg. According to our sketch, 30.4 would not be a very unusual mean value for a sample of 60 cars to obtain (it is not in the extreme tail of the distribution).

d. The CLT predicts a sampling distribution that would be normal with mean 31 mpg and standard deviation 3/= .2449 mpg. According to our sketch, 30.4 would be a somewhat unusual mean value for a sample of 150 cars to obtain.

e. Only in part (a) where we can’t calculate a probability there unless we know for such population distribution is normal. In the other cases the population shape does not matter because all of these sample sizes are at least 30, none of these responses depend on knowing the shape of the population distribution.

Activity 15-7ab (p. 303)

a. With n = 75 and p = .45, because 75(.45) = 33.75 and 75(.55) = 41.25 both exceed 10 and we are considering random samples, the CLT says the distribution of the sample proportions will be approximately normally distributed with mean = .45 and standard deviation =.

b. Axis label: sample proportion of candies that are orange

Reasonable guesses would be 10-20%

Activity 15-8abd

a. Now with n = 175, the CLT will still apply and claims that the distribution of sample proportions will be approximately normally distributed with mean = .45 and standard deviation =. The only change from when the sample size was 75 is the standard deviation – which is smaller now.

b.

Reasonable guesses would be 5-15%.

d. This probability is smaller than when the sample size is 75. This makes sense because the standard deviation (spread) has decreased and thus there are fewer sample proportions as far from the center of .45.

Activity 16-14 (p. 327-8)

a. A 95% confidence interval for the proportion of all viewers who favored Santos is:

Informal .54 + 2 = .54 + 2(.01434) = .54 + .0287 = (.511, .569)

Applet

b. Yes – since all the values in this interval are greater than .50, this suggests that more than ½ the population favored Santos.

Activity 16-15 (d)-(f)

d. If we were to repeat this procedure many, many times, always using random samples of 116 pages of Sports Illustrated (and 130 pages of Soap Opera Digest), 95% of the time we would create intervals that would contain the population proportion of the magazine’s pages that contain ads.

e. Yes – each interval contains the sample proportion of pages with ads

f. This question was silly because this value () is the center of the interval. We cannot create the interval without it.

Activity 17-13 ***

a. We can determine this: In the sample = 497/1309 = .38 considered themselves to be political moderates. This is clearly greater than 1/3.

b. We want to find the probability that more than 497 in a sample of 1309 consider themselves moderates, in other words Pr( > 497/1309 ) = .380 where we are told =.333

The CLT applies here because 1309(.333) = 217.9 > 10, 1309(.667) = 873.1 >10, and we have a random sample.

= 3.61 p-value = Pr(Z > 3.61) ≈ .0000

(you should realize this z-value corresponds to a very small probability even without using Table II or technology)

c. You expect the p-value to increase since the observed sample proportion (124/327 = .38) is the same but the sample size is smaller so there will be more random sampling variability. This implies the observed sample result is less surprising corresponding to a larger p-value.

d.

The p-value did indeed increase.

Activity 18-8 (p. 365)

a. = -6.8

b.

c. No – this significance test has no meaning because we know that women constitute less than half of the entire 2007 U.S. Senate since we have taken a census of the population of interest. We know they make up exactly 16% of the 2007 Senate.

Activity 19-17 (p. 389)

a. The observational units are the adult American who were interviewed by the GSS. The variable is the number of close friends that the adult American has. This variable is quantitative.

b. A t-interval is valid in spite of the strong right skew because the sample size is very large (1467).

c. 95% confidence interval for , the mean number of number friends in the population of all adult Americans.

Informal 1.987 + 2(1.7708/√1467) = 1.987 + .0924 = (1.894, 2.079)

Applet

We are 95% confident that the mean number of close friends in the population of American adults is between 1.894 and 2.079 friends.

d. The reasonable interpretations of this interval are:

· You can be 90% confident that the mean number of close friends in the population is between the endpoints of this interval.

· If you repeatedly took random samples of 1467 people and constructed t-intervals in this same manner, 90% of the intervals in the long run would include the population mean number of close friends.

e. “Ninety percent of all people in this sample reported a number of close friends within this interval” is incorrect because that is not what 90% confidence refers to. See either of the statements above for a correct interpretation.

“If you took another sample of 1467 people, there is a 90% chance that its sample mean would fall within this interval” is incorrect because we are not trying to capture sample means – we are trying to capture the population mean.

“If you repeatedly took random samples of 1467 people, this interval would contain 90% of your sample means in the long run” is not a correct interpretation because the interval would definitely contain the one sample mean you used to create the interval. We cannot predict how many of the other sample means it would contain – the interval procedure is estimating the population mean. We are not saying other sample means should be within 2 standard deviations of the one we observed, but that sample means in general should fall within 2 standard deviations of the actual population mean.

It is incorrect to say “this interval captures the number of close friends for 90% of the people in the population” because this interval estimates the mean number of friends – not the number of individual friends for any person.

f. If the sample size were large, the interval would have the same midpoint, but would be narrower.

If the sample mean were larger the interval would have a larger midpoint, but would have the same width.

If the sample values were less spread out, the standard deviation would be smaller, so the margin-of-error would be smaller, so interval would be narrower (but would have the same midpoint).

If every person in the sample reported one more close friend, the sample mean would be larger (by 1), so the midpoint of the interval would increase by 1, but the width would be unchanged.

Activity 20-8 (p. 404)

a. In this context represents the average IQ of all people who claim to have had an intense experience with an UFO.

b. This is a one-sided test, because we wish to test whether or not the average IQ of this group is greater than 100.

c. In order for this procedure to be valid, we need to know that the population of IQ scores for this group is approximately normally distributed, because the sample size is small (n = 25 < 30). We also need to be willing to believe that this sample is representative of all individuals who claim to have had such an experience.

d.

= 0.90

e. From the graph, we can ballpark the p-value between .1 and .33 (actual value .19). Based on the t value being less than 1 and the large amount shaded on the graph, this is not a surprising outcome when the population mean equals 100.

f. If the average IQ of this group is really 100, then we could expect to see a random sample of 25 people from this group with an average IQ of at least 101.6 in about 19% of samples by random chance alone. Since this would be a fairly common occurrence, we have no reason to doubt that the mean IQ of the population (those who claim to have had an intense experience with an UFO) is 100.

Activity 21-13

a.

This bargraph indicates that mothers given the placebo were about 3 times as more likely to have babies that were HIV positive than were the mothers given AZT.

b. i) The data are from randomly assigning subjects to two treatment groups. This condition is met.

ii) The number of successes and failures in each group should be at least 5. This condition is also met.

c. The null hypothesis is that AZT and a placebo are equally effective in reducing mother-to-infant transmission of AIDS. Specifically, the proportion of HIV-positive babies born to mothers who could potentially take AZT is the same as the proportion of HIV-positive babies born to mothers who could potentially take a placebo. In symbols, the null hypothesis is H₀: π_AZT = π_placebo.

The alternative hypothesis is that AZT is more effective than a placebo for reducing mother-to-infant transmission of AIDS, or that the proportion of HIV-positive babies born to mothers who could potentially take AZT is smaller than the proportion of HIV-positive babies born to mothers who could potentially take a placebo. In symbols, the alternative hypothesis is H_a: π_AZT < π_placebo.

With such a small p-value, reject H₀ at the α = .01 significance level.

We have very strong statistical evidence that AZT is more effective than a placebo for reducing mother-to-infant transmission of AIDS.

d.

We are 99% confident the difference in HIV transmission rates is between 5.33 and 23.95 percentage points. Since the values in our interval are all negative, we know that the AZT transmission rate is lower than the placebo transmission rate by somewhere between 5.33 to 23.95 percentage points.

e. Since this was a well-designed experiment, we can conclude that AZT caused the observed difference in HIV transmission rates. If AZT and a placebo were equally effecting in reducing mother-to-infant transmission of AIDS, we virtually never see sample results as or more extreme as those we saw in this experiment by random assignment alone. We are 99% confident in concluding that AZT lowers the HIV transmission rate somewhere between 5.33 and 23.95 percentage points over that of a placebo.

Activity 22-5 (p. 451)

From Activity 22-1: male n = 654, = 1.861, s = 1.777; female n = 813, = 2.089, s = 1.760

Note: the difference in sample means (female – male) = 2.089 - 1.861 = .228

a. If all of the women sampled had one more friend, the mean for the women would increase by one, and all other statistics would remain the same. This would increase the size of the difference in the two means (new difference = 1.228) and therefore would also increase the absolute value of the test statistic (the difference between the sample means would increase but the denominator would not change), and thus would decrease the p-value.

Note: Since we are a two-sided alternative, it doesn’t matter which direction we subtract in to find the p-value

b. If all of the men sampled had one more friend, the mean for the men would increase by one, and all other statistics would remain the same. The absolute value of the new difference in means would be actually be larger (2.089 – 2.861 = -.772). This would increase the absolute value of the test statistic (the difference between the sample means would increase, but the denominator would not change), and thus would decrease the p-value.

c. If every man and every woman sampled had one more close friend than they originally reported, then the sample means for the men and for the women would increase by one. The sample standard deviations and sample sizes would not change. The difference between the sample means would not change, so the test statistic value would not change, and thus the p-value would not change.

d. If both sample standard deviations were larger, the denominator of the test statistic would be larger, and thus the test statistic would be smaller. This would make the p-value larger.

e. If both sample sizes were larger, the denominator of the test statistic would be smaller, and thus the test statistic would be larger (in absolute value). This would make the p-value smaller.

Activity 22-8 (p. 453)

a. age - comparison of means

weight – comparison of means

gender – comparison of proportions

number of cigarettes smoked – comparison of means

whether the person made a previous attempt to quit smoking – comparison of proportions

b. The researchers would hope to fail to reject the null hypotheses in these tests because the null hypotheses would be that the two groups (those that use the nicotine lozenge and those that don’t) are identical with regard to each of these background variables. The researchers would be hoping that they would not find a statistically significant difference between the groups on any of these variables so that a difference between the groups on the response variable could be attributed to the nicotine lozenge.