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Practice Final Exam Questions (2) -- Answers
Part A. Multiple Choice Questions. For each question, you are encouraged to give a reason or show work for partial credit. You must show your work or reason if the question is marked with an asterisk (*).
- Confidence intervals are useful when trying to estimate _______. a. unknown parameters ๏ง b. known parameters c. unknown statistics d. known statistics
- The one-sample z statistic is used instead of the one-sample t statistic when ______. a. ฮผ is known b. ฮผ is unknown c. ฯ is known ๏ง d. ฯ is unknown
- The ____ the P-value, the stronger the evidence against the null hypothesis provided by the data. a. larger b. smaller ๏ง
- (*) The test statistic for a two-sided significance test for a population mean is z = โ2.12. What is the corresponding P-value? a. 0. b. 0.034 ๏ง c. 0. d. 0. Answer: P(Z < -2.12) = 0.0170. Since this is a two-sided test, P-value is twice as much to include the other tail. 0.0170 * 2 = 0.034.
- The probability you reject the null hypothesis when in fact the null hypothesis is true is called __________. a. a Type I error ๏ง b. a Type II error c. the power
- (*) A random sample of 20 observations produced a sample mean of ๐ฅ๐ฅฬ
= 92.4 and s = 25.8. What is the value of the standard error of ๐ฅ๐ฅฬ
? a. 4. b. 15. c. 1. d. 5.8 ๏ง e. 2. Answer: ๐๐๐๐๐ฅ๐ฅฬ
= (^25) โ20.^8 = (^425). 472.^8 = 5.
- (*) The heights (in inches) of adult males in the United States are believed to be Normally distributed with mean ฮผ. The average height of a random sample of 25 American adult males is found to be ๐ฅ๐ฅ ฬ
= 69. inches, and the standard deviation of the 25 heights is found to be s = 4.15. A 90% confidence interval for ฮผ is a. 69.72 ยฑ 1. b. 69.72 ยฑ 1. c. 69.72 ยฑ 1.42 ๏ง
Answer: The t critical value for 90% CI for df = 24 is 1.711. So margin_or_error = (1.711) โ
- Suppose we were interested in determining if there were differences in the average prices among two local supermarkets. We randomly pick six items to compare at both supermarkets. Which statistical procedure would be best to use for this study? a. Matched-pairs t procedure ๏ง if you interpreted as โsame six itemsโ b. One-sample t test c. Two-sample t test ๏ง if you interpreted as โdifferent six itemsโ d. None of the above
- (*) Perform a one-sample t-test using the following statistics:
n = 5 ๐ฅ๐ฅฬ
= 3.871 s = 0.
The null hypothesis is ฮผ = 5.0 is a. accepted at the 5% level; accepted at the 1% level. b. accepted at the 5% level; rejected at the 1% level. c. rejected at the 5% level; accepted at the 1% level. ๏ง d. rejected at the 5% level; rejected at the 1% level.
Answer: Test statistic: ๐ก๐ก = ๐ฅ๐ฅฬ
โ๐๐๐ ๐
โ๐๐
= 3 .871โ5 0. 679.^0
โ
Assuming a two-sided test, for df = 4,
- 95% CI of the t-statistic assuming H0 is [-2.776, 2.776]
- 99% CI of the t-statistic assuming H0 is [-4.604, 4.604] Therefore we reject H0 for alpha = 0.05 but not for alpha 0.01.
- (*) You buy a package of 122 Smarties and 19 of them are red. What is a 95% confidence interval for the true proportion of red Smarties? a. (0.092, 0.220) ๏ง b. (0.103, 0.230) c. (0.085, 0.199)
Answer: ๐๐ฬ = 12219 = 0.156. margin_of_error = (1.96)^ โ ๏ฟฝ(^0.^156122 )โ(^0.^844 )= (1.96)^ โ (0.033) = 0.0647. So the 95% CI is [0.156 โ 0.0647, 0.156 + 0.0647], which yields [0.0913, 0.2207].
Answer: If the alternative hypothesis was Ha: ฮผ 1 - ฮผ 2 < 0, we can reject the null hypothesis at alpha = 0.05 level because the p-value would be between 0.025 and 0.05.
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Part C. Problems
- One-sample t-test. To test the hypothesis that eating fish makes one smarter, a random sample of 12 persons take a fish oil supplement for one year and then are given an IQ test. Here are the results:
116 111 101 120 99 94 106 115 107 101 110 92
Test using the following hypotheses, report the test statistic with the P-value, then summarize your conclusion.
H0: ฮผ = 100 Ha: ฮผ > 100
Answer:
Hypotheses:
H0: ฮผ = 100 (no effect -- eating fish does not help increase the mean IQ) Ha: ฮผ > 100 (effect -- eating fish helps increase the mean IQ)
Test statistic:
From the data, we obtain ๐๐๏ฟฝ = ๐๐๐๐๐๐ and ๐ฌ๐ฌ๐๐ = ๐๐. ๐๐๐๐. Then we get
๐๐ =
๐๐๏ฟฝ โ ๐๐๐๐ ๐๐๐๐ โ๐๐
=
๐๐๐๐๐๐ โ ๐๐๐๐๐๐ ๐๐. ๐๐๐๐ โ๐๐๐๐
=
๐๐ ๐๐. ๐๐๐๐ =^ ๐๐.^ ๐๐๐๐.
P-value:
Because n = 12, we use the t distribution with df = 11 to find the probability. According to Table D, for t = 2.35 for df = 11, the probability is between 0.01 and 0.02. Since this is a one-sided (upper-tail) test, P-value is between 0.01 and 0..
Conclusion:
Since values between 0.01 and 0.02 are < 0.05, we reject the null hypothesis at the significance level 0.05 , and conclude that the fish oil supplement did make a significant increase in the mean IQ.
- Matched one-sample t-test. The water diet requires you to drink 2 cups of water every half hour from when you get up until you go to bed but eat anything you want. Four adult volunteers agreed to test this diet. They are weighed prior to beginning the diet and 6 weeks after. Their weights in pounds are
Person 1 2 3 4 mean _ s.d. Weight before 180 125 240 150 173.75 49. Weight after 170 130 215 152 166.75 36. Difference 10 -5 25 -2 7 13.
Conduct a one-sample t-test using the difference with the following hypotheses:
H0: Diff = 0 Ha: Diff โ 0
Report the test statistic with the P-value, then summarize your conclusion.
Answer:
Hypotheses:
H0: Diff = 0 (no difference -- there is no difference in the mean of the weight difference) Ha: Diff โ 0 (difference โ diet made difference in the mean of the weight difference)
Test statistic:
From the data, we know ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ^ = ๐๐ and ๐ฌ๐ฌ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ = ๐๐๐๐. ๐๐๐๐. Then we get
๐๐ = ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐๐๐๐ ๐๐๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ โ๐๐
= (^) ๐๐๐๐ โ๐๐. ๐๐^ ๐๐๐๐ โ๐๐
= (^) ๐๐.๐๐ ๐๐๐๐ = ๐๐. ๐๐๐๐๐๐.
P-value:
Because n = 4, we use the t distribution with df = 3 to find the probability. According to Table D, for t = 1.026 for df = 3, the probability is between 0.15 and 0.20. Since this is a two-sided test, P-value is between 0.30 and 0..
Conclusion:
Since values between 0.30 and 0.40 are > 0.05, we fail to reject the null hypothesis at the 0.05 significance level. We do not have enough evidence to conclude that the water diet has an impact on the weight.
- Two-sample t procedure. Two different alloys are being considered for making lead-free solder used in the wave soldering process for printed circuit boards. A crucial characteristic of solder is its melting point, which is known to follow a Normal distribution. A study was conducted using a random sample of 21 pieces of solder made from each of the two alloys. In each sample, the temperature at which each of the 21 pieces melted was determined. The mean and standard deviation of the sample for Alloy 1 were x^1 = 218.9ยบC and s^1 = 2.7ยบC; for Alloy 2 the results were x^2 = 215.5ยบC and s^2 = 3.6ยบC. If we were to test H 0 : ฮผ 1 =ฮผ (^2) against H a: ฮผ 1 โ ฮผ (^2) , what would be the value of the test statistic??
