Statistics in a Modern World 800
Solutions to Exam 4
- (c) Since zero is well-contained within the interval, there is no
statistical evidence of a difference.
- (a) The U.S. interval (41,136) is much higher than the Australia
interval (9,36) with no overlap.
- (a) yes!
-
- (ii) z test about a mean (null hypothesis says mean IQ of children
of women smokers equals 100)
- (iv) chi-squared test for a relationship between two categorical
variables (conviction or acquittal, and religion)
- (iii) two-sample z test (comparing 2 means)
- (i) z test about a proportion (null hypothesis says proportion correct
equals .25)
- (c) P-value not small at all indicates no evidence of a relationship
-
- (ii) approximately the same: both should equal population proportion
- (iii) more for teaspoons: more spread for smaller samples
- (iii) more normal for Tablespoons: Central Limit Theorem gave this
result
-
- mean of sample means=population mean=3.5
- standard deviation of sample means = population s.d. over square root
of sample size: 1.7/5=.34
- (iii) approximately normal because 25 is large enough (Central Limit
Theorem again)
- within 1 standard error of mean: 3.5 plus or minus .34 = (3.16,3.84)
-
- .27 plus or minus 2 times square root of (.27)(1-.27)/294)
=.27 plus or minus .05, or (.22,.32)
- yes, it's a minority because the above interval is so far below .5.
- (i) is the only correct interpretation
- 2(.04)=.08
-
- (ii) fail to reject the null hypothesis, even when it is false:
small sample leads to small z which leads to large P-value which does not
provide evidence to reject
- (i) find a difference to be statistically significant when it has
little practical significance: large sample leads to large z which leads to
small P-value which provides evidence to reject
-
- (v) null hypothesis: population mean for women = .47
- (iv) alternative hypothesis: population mean for women less than .47
- (29-47)/(16/10)=-11.25
- P-value is approximately zero since the z score is so far below zero
- (i) yes: small P-value tells us to reject the null hypothesis and
conclude the alternative is true.
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