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1. A Random sample of monthly rent paid by 12 college seniors living off campus gave the results below (in dollars). Find a 99 percent confidence interval for µ , assuming that the same is from a normal population.
900 810 770 860 850 790 810 800 890 720 910 640
Sample mean = 812.5, standard deviation = 78.407
For 99% confidence, z = 2.5758
Margin of error = 2.5758*78.407/sqrt(12) = 58.3
Lower limit = 812.5 - 58.3 = 754.2
Upper limit = 812.5 + 58.3 = 870.8
So, 99% confidence interval for µ is (754.2, 870.8)
2. Of 43 bank customers depositing a check, 18 received some cash back. (a) construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (b) check the normality assumption.
a) p = 18/43 = 0.4186
For 90% confidence, z = 1.6449
Margin of error = 1.6449*sqrt(0.4186*(1-0.4186)/43) = 0.1237
Lower limit = 0.4186 - 0.1237 = 0.2949
Upper limit = 0.4186 + 0.1237 = 0.5423
So, 90% confidence interval for proportion is (0.2949, 0.5423).
b) np = 0.4186*43 = 18
n(1-p) = 43*(1-0.4186) = 25
Since both values are more than 5, normality assumption is valid.
3. The recent default rate of all student loans is 5.2 percent. In a recent random sample of 300 loans at private universities there were 9 defaults. (a) does this sample show sufficient evidence that the private university loan default rate is below the rate for all universities, using a left-tailed test at α=.01? (b) calculate the p-value (c) verify that the assumption of normality is justified.
a) H0: p >= 0.052
H1: p < 0.052
p = 9/300 = 0.03
z-statistic = (0.03-0.052)/sqrt(0.052*(1-0.052)/300) = -1.7162
Critical value = -2.3263
Since test statistic is greater than critical value, we fail to reject null hypothesis.
There isn’t sufficient evidence that the private university loan default rate is below the rate for all universities.
b) p-value = 0.043
c) np = 0.03*300 = 9
n(1-p) = (1-0.03)*300 = 297
Since both values are more than 5, normality assumption is valid.
4. Find the p-value for each test statistic.
a. right-tailed test, z =+1.34 b. left-tailed test, z=-2.07 c. two-tailed test, z=-1.69
a) 0.0901
b) 0.0192
c) 0.091
5. the manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a) at α =.05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer’s claim is over stated? (b) why might the assumption of a normal population be doubtful?
a) H0: µ = 530
µ < 530
t = (510-530)/(50/sqrt(16)) = -1.6
degrees of freedom = 16-1 = 15
p-value for left tailed test = 0.0652
Since p-value is more than 0.05, H0 can’t be rejected.
We can’t say manufacture’s claim is overstated.
b) Assumption of normality is doubtful because more scanning would be done per hour during flight times, than during idle times.
6. A new drug is being tested to see if reduces the number of migraine headaches. Assuming equal variances, at a α=.025, do these samples of number of monthly migraines from eight volunteers who are taking the drug and eight who are not show a significant reduction in the mean number of monthly headaches?
Topiramate: 3 3 2 4 3 4 3 5
Control: 3 3 5 5 7 5 5 4
Hint: use excel data analysis T test assuming equal variances
H0: µ1 = µ2
H1: µ1 < µ2
Excel output
|t-Test: Two-Sample Assuming Equal Variances |
| | | |
| |Topiramate |Control |
|Mean |3.375 |4.625 |
|Variance |0.8393 |1.6964 |
|Observations |8 |8 |
|Pooled Variance |1.2679 | |
|Hypothesized Mean Difference |0 | |
|df |14 | |
|t Stat |-2.2203 | |
|P(T ................
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