Workshop Statistics: Discovery with Data and Fathom

Topic 23: More Inference Considerations

Activity 23-1: Racquet Spinning (cont.)

(a) (.362, .558)
(b) yes
(c) Based on our answer to (b), we would not expect a significance test of whether q differs from .5 to be significant at the .05 level because the interval indicates that .5 is a plausible value for q.
(d) test statistic = -.8
     p-value = .424
Since the p-value is so large, the sample proportion does not differ significantly from one-half at the .05 significance level.  So, yes, our expectation in (c) was realized.
(b)-(h)
Hypothesized value
Contained in 95% CI?
Test statistic
p-value
Significant at .05?
.35
no
2.31
.021
yes
.40
yes
1.22
.221
no
.50
yes
-.80
.424
no
.55
yes
-1.81
.070
no
.60
no
-2.86
.004
yes

(i) If the hypothesized value is contained in the 95% confidence interval, it will not be significant at the .05 level, and vice versa.
 

Activity 23-2: Racquet Spinning (cont.)

(a) sample proportion: .565;  test statistic: 1.84;  p-value: .066;  significant at .05?: no
(b) sample proportion: .575;  test statistic: 2.12;  p-value: .034;  significant at .05?: yes
(c) sample proportion: .650;  test statistic: 4.24;  p-value: .000; significant at .05?: yes
(d) a and b
(e) b and c
 

Activity 23-3: Cat Households (cont.)

(a) 99.9% confidence interval: (.268, .278);  Hypotheses: Ho: q = .25,  Ha: q > .25;  test statistic: 15.02;  p-value: less than .0002 (.0000 to many decimal places).
(b) Yes, because the entire 99.9% confidence interval is greater than .25, and the test of significance reveals very strong evidence against the null hypothesis that theta equals .25.
(c) No, it is most likely only 1-2% more than 25% since the 99.9% confidence interval is only (.268, .278).
(d) confidence interval
 

Activity 23-4: Hypothetical Baseball Improvements

Students' answers to (a)-(g) may differ.  These are meant to be sample answers.
(a)

         <------------number of hits -------------->
        This distribution is symmetrical, with the center at 7.  The spread is from 1 to 15.
(b) About 13, because 5% of the data points lie above this point.
(c)


       <------------number of hits ---------------->
 
   The top distribution is using a .250 hitter, while the bottom is using a .333 hitter.  This distribution for the .333 hitter is also symmetrical, with the center at 10.  The spread is from 3 to 18.  There is quite a bit of overlap between the two distributions.

(d) The .333 hitter exceeded 13 hits about 12 percent of the time.
(e) No.  Even though he is actually a .333 hitter, there is only about a 12% chance that he will get enough hits (>13) to convince us that he's not a .250 hitter.  Also, we know that since the overlap between the two distributions is so great, most of the time a .333 hitter will have a number of hits that could easily fall on the .250 hitter's distribution, meaning it would bot be significantly better than a .250 hitter.
(f) Approximate power = .12
(g)


 
 
    There is less overlap between the two distributions.  From the first distribution (that of a .250 hitter), he'd be in the top 5% of performances if he got more than 34 hits or so. When he is a .333 hitter, he will get more than 34 hits almost 50% of the time. Thus, there is a much higher chance that he will be able to convince us he is better than a .250 hitter. Increasing the sample size gives us more evidence and increases the power we have to detect that his performance has improved.
(h) more powerful (the distribution will now be centered higher and have less overlap with .250 distribution), he's a much better hitter so it's much easier for him to perform convincingly higher than a .250 hitter.
(i) more powerful. He doesn't need to perform as high to convince us he has improved.
(j) alternative value, level of significance

Activity 23-5: Halloween Practices (cont.)

(a)1.96sqrt(.69(.31)/1005)= .0286
(b) We would need a larger sample because larger samples make the statistic more accurate.
(c) Solve 1.96sqrt(.69(.31)/n)=.01 for n to find that n=8,218
(d) We would need even more people because increasing the confidence without changing the margin or error would require a larger sample size.
(e) Solve 2.576sqrt(.69(.31)/1005) to find n=14,194
(f) The population size did not enter into these calculations at all.  The answers to (c) and (e) would be no different if the population of interest were all California adults rather than all American adults.
(g) Every person in the population would have to be interviewed to determine the value of the population proportion exactly, with 100% confidence.
 

Activity 23-6: Hypothetical ATM Withdrawals (cont.)

(a)
 
Sample size
Sample mean
Sample std. dev.
95% confidence interval for m
machine 1
50
70
30.3
(61.39, 78.61)
machine 2
50
70
30.3
(61.39, 78.61)
machine 3
50
70
30.3
(61.39, 78.61)

(b)



        These distributions are very different from each other.  While many of their descriptive statistics, such as sample size, mean, and standard deviation are the same, the distributions still differ greatly.
 

Activity 23-7: Female Senators (cont.)

(a)( .034, .146)
(b) no
(c) The technical conditions necessary for this procedure to be valid are not fully met.  This sample is not a simple random sample from the population of interest.  The male/female ratio in the 1999 U.S. Senate is not representative of all humans.
(d) The interval does not make sense for this purpose because we know the population proportion of women to be .09 for the 1999 U.S. Senate.