Download STAT 200 FINAL EXAM 1 ; Latest Complete Solutions with shown working, University Of Maryla and more Exams Statistics in PDF only on Docsity!
1. True or False. Justify for full credit. (a) If the variance of a data set is zero, then all the observations in this data set are zero. No, all the observations will be same but it is not necessary that all the observations will be zero, because if all the observations are same even then the variance will be zero. (b) If P(A) = 0.4 , P(B) = 0.5, and A and B are disjoint, then P(A AND B) = 0.9. No, because if A and B are disjoint then P(A AND B) = 0. (c) Assume X follows a continuous distribution which is symmetric about 0. If , then. Yes, P(X<-3) will be less than 0.3 because the value will be towards negative side. (d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter. Yes, a 95% confidence interval is wider than a 90% confidence interval of the same parameter because increasing the confidence limit we increases the acceptance region and reduces the rejection region. (e) In a right-tailed test, the value of the test statistic is 1.5. If we know the test statistic follows a Student’s t-distribution with P(T < 1.5) = 0.96, then we fail to reject the null hypothesis at 0.05 level of significance. No, we will not be able to reject the null hypothesis as we reject the null hypothesis is test statistic is greater than the tabulated value of t.
Refer to the following frequency distribution for Questions 2, 3, 4, and 5. Show all work. Just the answer, without supporting work, will receive no credit. The frequency distribution below shows the distribution for checkout time (in minutes) in UMUC MiniMart between 3:00 and 4:00 PM on a Friday afternoon. Checkout Time (in minutes) Frequency Relative Frequency 1.0 - 1.9 3 2.0 - 2.9 12 3.0 - 3.9 0. 4.0 - 4.9 3 5.0 - 5. Total 25
2. Complete the frequency table with frequency and relative frequency. Express the relative frequency to two decimal places. Checkout Time (in minutes) Frequency Relative Frequency 1.0 - 1.9 3 0. 2.0 - 2.9 12 0. 3.0 - 3.9 5 0. 4.0 - 4.9 3 0. 5.0 - 5.9 2 0. Total 25 1 3. What percentage of the checkout times was at least 3 minutes? Percentage of the checkout times was at least 3 minutes = (5 + 3 +2)/25* = 40% 4. In what class interval must the median lie? Explain your answer. We will make the intervals continuous by subtracting 0.05 from lower limit and adding 0.05 in upper limit therefore the corresponding intervals will be 0.95-1.95, 1.95 - 2.95, 2.95 - 3.95, 3.95 - 4.95, 4.95 - 5.95. N=25 therefore, N/2=12.5, the immediate less observation lies in 2.0-2.9 i.e. 1.95-2.95. 5. Does this distribution have positive skew or negative skew? Why?
Refer to the following information for Questions 6 and 7. Show all work. Just the answer, without supporting work, will receive no credit. Consider selecting one card at a time from a 52-card deck. (Note: There are 4 aces in a deck of cards)
6. If the card selection is without replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form) (5 pts) The probability that the first card is an ace and the second card is also an ace: P(First card is an Ace and Second card is also an Ace)=(4/52)(3/51) =0. 7. If the card selection is with replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form) The probability that the first card is an ace and the second card is also an ace: P(First card is an Ace and Second card is also an Ace)=(4/52)(4/52) = 0.
Refer to the following situation for Questions 8, 9, and 10. The five-number summary below shows the grade distribution of two STAT 200 quizzes for a sample of 500 students. Minimum Q1 Median Q3 Maximum Quiz 1 15 45 55 85 100 Quiz 2 20 35 50 90 100 For each question, give your answer as one of the following: (a) Quiz 1; (b) Quiz 2; (c) Both quizzes have the same value requested; (d) It is impossible to tell using only the given information. Then explain your answer in each case.
8. Which quiz has less interquartile range in grade distribution? Quiz 1 Because the inter quartile range i.e. Q3-Q1=40 for quiz 1 is less than the interquartile range Q3- Q1= 55 for quiz 2. 9. Which quiz has the greater percentage of students with grades 90 and over? Quiz 2 Because 75 th^ percentile of quiz 2 is having grades 90 and this is not the case for quiz 1. 10. Which quiz has a greater percentage of students with grades less than 60? Quiz 1 Because median of quiz 1 is having grades 55 and this is not the case for quiz 1.
14. A combination lock uses three distinctive numbers between 0 and 49 inclusive. How many different ways can a sequence of three numbers be selected? ( Show work ) To find the different ways in which a sequence of three numbers can be selected is:
n Cr = 50 C 3
15. Let random variable x represent the number of heads when a fair coin is tossed three times. Show all work. Just the answer, without supporting work, will receive no credit. (a) Construct a table describing the probability distribution. Number of Heads (X) 0 1 2 3 Probability [P(X=r)] 0.125 0.375 0.375 0. (b) Determine the mean and standard deviation of x. (Round the answer to two decimal places) Mean= Expected Value of x = ∑xp(x) = 00.125 + 10.375 + 20.375 + 30. =1. Standard Deviation= sqrt(∑xxp(x)- (∑xp(x))^2) =000.125 + 110.375 + 220.375 + 330.125 – (1.5)^ =0.
Refer to the following information for Questions 17, 18, and 19. Show all work. Just the answer, without supporting work, will receive no credit. The lengths of mature jalapeño fruits are normally distributed with a mean of 3 inches and a standard deviation of 1 inch.
17. What is the probability that a randomly selected mature jalapeño fruit is between 1.5 and 4 inches long? (5 pts) Here we have μ = 3, σ = 1 Z score can be calculated as Z=(X-μ)/(σ) We need to find P(1.5<X<4) = P((1.5-3)/(1)<Z<(4 - 3)/(1))=P(-1.5<Z<1)= P(Z<1)- P(Z<-1.5) = 0.8413- 0.0668= 0. The probability that a andomly selected mature jalapeño fruit is between 1.5 and 4 inches long is 0.7745. 18. Find the 90 th^ percentile of the jalapeño fruit length distribution. We need to find X (X- μ)/ σ (X-3)/1= 1. X = 1.2816*1+3 = 4. The 80th percentile of the pecan tree height distribution is 11. 19. If a random sample of 100 mature jalapeño fruits is selected, what is the standard deviation of the sample mean? The standard deviation of the sample mean will be (σ/sqrt(n)) = 1/sqrt(100) = 1/10 = 0.
20. A random sample of 100 light bulbs has a mean lifetime of 3000 hours. Assume that the population standard deviation of the lifetime is 500 hours. Construct a 95% confidence interval estimate of the mean lifetime. Show all work. Just the answer, without supporting work, will receive no credit. 95% confidence interval is given as Sample Mean ± Z(critical value)SE Sample mean = 3000, n = SE = standard deviation/sqrt(n)=500/sqrt(100)=500/10 = 50 Z(critical value at 0.05)=1. So the lower limit is LL = 3000-1.9650 = 2902 UL = 3000+1.96*50 = 3098 So 95% confidence interval estimate of the mean lifetime is (2902, 3098)
22. Consumption of large amounts of alcohol is known to increase reaction time. To investigate the effects of small amounts of alcohol, reaction time was recorded for five individuals before and after the consumption of 2 ounces of alcohol. Do the data below suggest that consumption of 2 ounces of alcohol increases mean reaction time? Reaction Time (seconds) Subject Before After 1 6 7 2 8 8 3 4 6 4 7 8 5 9 8 Assume we want to use a 0.01 significance level to test the claim. (a) Identify the null hypothesis and the alternative hypothesis. H 0 : H 1 :
D ´^ = 0
D ´^ ≠ 0
Reaction Time (seconds) Subject Before After d (d-d )¯ ^ 1 6 7 1 0. 2 8 8 0 0. 3 4 6 2 1. 4 7 8 1 0. 5 9 8 - 1 2. 0.6 5. (b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. The test statistic is defined as:
d ´
t= ¿ -^ D
/sqrt [ (Σ(di - d
)^2 / (n - 1) )/n] under H0, the test statistic is: t=(0.6-0)/sqrt(5.2/(5-1)/5) =1.
(c) Determine the P - value. Show all work; writing the correct P-value, without supporting work, will receive no credit. P-value: p-value= P(t < - 1.176696811) + P(t > 1.176696811) =0.15228061+0. =0. (d) Is there sufficient evidence to support the claim that consumption of 2 ounces of alcohol increases mean reaction time? Justify your conclusion. No, there is sufficient evidence to support the claim that consumption of 2 ounces of alcohol increases mean reaction time, because the p-value is greater than the level of significance 0.01 which tells us that we can accept the null hypothesis.
(d) Is there sufficient evidence to support the manager’s claim that the four types are equally popular? Justify your answer. No, there is no sufficient evidence support that the manager’s claim that the four types are equally popular because we are rejecting the null the null hypothesis and accepting the alternative which states that data is not consistent with the four different types of Halloween candy bags
24. A random sample of 4 professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars). x 0 1 3 5 y 1 2 3 8 (a) Find an equation of the least squares regression line. Show all work; writing the correct equation, without supporting work, will receive no credit. The equation of the least squares regression line is: y=b0 + b1*x where b1 is the slope and b0 is the intercept (the point where the line crosses the y axis) y x xy x^ 1 0 0 0 2 1 2 1 3 3 9 9 8 5 40 25 Total 14 9 51 35 b0 =
∑ y^ ∑ x
∑ x^ ∑ x y
n ∑ x^2 −(∑ x )
4 ∗ 35 −( 9 )^2 =
59 =^ 0.
b1 =
n ∑ xy −∑ x ∑ y
n ∑ x^2 −(∑ x )
4 ∗ 35 −( 9 )^2 =
Therefore, the regression line equation is : y=0.525424 + 1.322034x (b) Based on the equation from part (a), what is the predicted value of y if x = 4? Show all work and justify your answer. The predicted value of y if x = 4 is y=0.525424 + 1.322034 =5.
