Homework 09 - Answers and Points

# Homework 09 - Answers and Points -  ...

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Unformatted text preview:   Homework Assignment 9  Psychology 60  Spring 2010    ‐   For homework problems requiring calculation, you must show all of your work,  including intermediate steps, in order to receive credit.        1) (4 points total) A researcher has constructed an 80% confidence interval from sample data,  and found the interval to be μ = 45 ± 8, using a sample of n = 25 scores.   a. (1 point) What would happen to the width of the interval if the researcher used a  larger sample size (assuming other factors are held constant)?  The width of this interval would decrease (the standard error and the tconf would be  smaller leading to a smaller interval)    b. (1 point)What would happen to the width of the interval if the researcher  constructed a 90% confidence interval rather than an 80% confidence interval?   The width of this interval would increase (the tconf for a 90% interval is larger than  the tconf for an 80% interval)    c. (1 point)What would happen to the width of the interval if the sample variance  increased (assuming other factors are held constant)?  The width of this interval would increase (the standard error would be larger leading  to a larger interval)    d. (1 point)What would happen to the width of the interval if the mean of the sample  were twice as large (assuming other factors are held constant)?  The width of this interval would be unaffected (neither the standard error nor the  tconf would change as a result of the mean changing)        2) (12 points total) A researcher obtains a sample from an unknown population and computes  a sample mean of  X = 43 , and a standard deviation of s = 6.   a. (3 points) If the sample has n = 16 scores, what is the 80% confidence interval to  estimate the unknown population mean.  sX = s2 62 36 , sX = , sX = , s X = 1.5   n 16 16   80% confidence interval with df = 15: t conf = 1.341     µ = X ± t conf × s X , µ = 43 ± 1.341 × 1.5, µ = 43 ± 2.0115,      40.9885 < µ < 45.0115    b. (3 points) If the sample has n = 36 scores, what is the 80% confidence interval to  estimate the unknown population mean.     sX = s2 , sX = n 62 ,s = 36 X 36 , s = 1  36 X   80% confidence interval with df = 35 (actually df=30): t conf = 1.310     µ = X ± t conf × s X , µ = 43 ± 1.310 × 1.0, µ = 43 ± 1.310, 41.69 < µ < 44.31       c. (3 points) If the sample has n = 16 scores, what is the 95% confidence interval to  estimate the unknown population mean.     sX = s2 62 36 , sX = , sX = , s X = 1.5   n 16 16   95% confidence interval with df = 15: t conf = 2.131     µ = X ± t conf × s X , µ = 43 ± 2.131 × 1.5, µ = 43 ± 3.1965, 39.8035 < µ < 46.1965       d. (3 points) If the sample has n = 36 scores, what is the 95% confidence interval to  estimate the unknown population mean.     sX = s2 , sX = n 62 ,s = 36 X 36 , s = 1  36 X   95% confidence interval with df = 35 (actually df=30): t conf = 2.042     µ = X ± t conf × s X , µ = 43 ± 2.042 × 1.0, µ = 43 ± 2.042, 40.958 < µ < 45.042     3) (5 points total) Standardized measures seem to indicate that the average level of anxiety  has increased gradually over the past 50 years. In the 1950s, the average score on the Child  Manifest Anxiety Scale was μ = 15.1. Suppose a sample of n = 16 children today produces a  mean score of  X = 23.3 , and SS = 240.   a. (1 point) Based on the sample, make a point estimate of the population mean anxiety  score for children today.   The best point estimate is the observed sample mean, so the best point est is μ = 23.3     b. (4 points) Construct a 90% confidence interval estimating the population mean of  children today    s2 = SS 2 240 2 ,s= , s = 16                s X = df 15 s2 16 , sX = , s = 1  n 16 X   90% confidence interval with df = 15: t conf = 1.753     µ = X ± t conf × s X , µ = 23.3 ± 1.753 × 1.0, µ = 23.3 ± 1.753, 21.547 < µ < 25.053     4) (7 points total) A researcher would like to determine how physical endurance is affected by  a common herbal supplement. The researcher measures endurance for a sample of n = 9  participants. Each individual is then given a 30‐day‐suppy of herbs and, 1 month later,  endurance is measured again.  For this sample, endurance increased by an average of  X D = 6   with SSD = 216.    a. (1 point) Based on the sample, make a point estimate of the population mean  difference in endurance for individuals taking the herbs.     The best point estimate is the observed sample mean, so the best point est is μD = 6    b. (4 points) Construct a 95% confidence interval estimating the population mean  difference in endurance for individuals taking the herbs    2 sD = SSD 216 2 , s2 = , sD = 27            s XD = df 8 2 sD , sXD = n 27 , s XD = 3, s XD = 1.732   9   95% confidence interval with df = 8: t conf = 2.306     µ D = ( X D ) ± t conf × s XD , µ D = (6 ) ± 2.306 × 1.732, µ D = (6 ) ± 3.994, 2.006 < µ D < 9.994     c. (2 points) Describe what this confidence interval implies about whether the herbs  have an effect on endurance  Because zero is not contained in the 95% confidence interval, we have (at least) 95%  confidence that the true population mean difference score is not 0, implying that  there is some effect of herbs. In fact, we have 95% confidence the true mean  difference score is between about 2 and 10, so it is within that range that we believe  the true value of μD to be.       5) (11 points total) A developmental psychologist wants to determine whether infants display  any color preferences. A stimulus consisting of four color patches (red, green, blue and  yellow) is projected onto the ceiling above a crib. Instants are placed in the crib, on at a time,  and the psychologist records how much time each infant spends looking at each of the four  colors. The color that receives the most attention during a 100‐second test period is  identified as the preferred color that infant. The preferred colors for a sample of 60 infants  are shown in the following table:    Red  Blue  Yellow  12    Green  20  10  18  a. (2 points) What are the null and alternative hypotheses for this experiment?  H0: There is no preference in the population for red, green, blue or yellow    or H0: The percent of preference in the population for these colors is each 25%    H1: There is some preference in the population for red, green, blue or yellow    or H1: The percent of preference in the population for these colors is NOT each 25%      b. (2 points) What is the critical value of the relevant test‐statistic?    2 df = (c − 1), df = ( 4 − 1), df = 3, for α = 0.05 χ critical = 7.81       c. (2 points) What are the expected values for each of the colors in this experiment?    Expected values = n/c    Red  Green  Blue  Yellow  60/4 = 15  60/4 = 15  60/4 = 15  60/4 = 15        d. (5 points) Do the data indicate any statistically significant preference among the four  colors? Test using  α = 0.05  and explain what this means for the original research  question.    ⎛ ( f o − f e )2 ⎞ 2 χ observed = ∑ ⎜ ⎟ fe ⎝ ⎠ ⎛ (12 − 15 )2 ⎞ ⎛ (20 − 15 )2 ⎞ ⎛ (10 − 15 )2 ⎞ ⎛ (18 − 15 )2 ⎞ 2 χ observed = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ 15 15 15 15 ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎛ 9 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 9 ⎞ 2 χ observed = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 15 ⎠ 2 χ observed = 0.6 + 1.66 + 1.66 + 0.6 2 χ observed = 4.533         2 2 χ observed = 4.533, χ critical = 7.81   2 2 χ observed < χ critical , Fail to reject H 0 The data do not provide evidence for any color preference for infants at this age.    6) (12 points total) A local county is considering a budget proposal that would allocate extra  funding toward the renovation of city parks. A survey is conducted to measure public  opinion concerning the proposal. A total of 150 individuals responded to the survey: 50 who  live within the city limits, and 100 who live in the surrounding suburbs. The local county  wants to determine if there is any difference in preference for people who live in the city and  people who live in the surrounding suburbs. The frequencies of responses are as follows:    Opinion    Favor  Oppose  City  35 People  15 People  Suburb  55 People  45 People    a. (2 points) What are the null and alternative hypotheses for this study?    H0: Region living and opinion on funding are independent     H1: Region living and opinion on funding are not independent      b. (2 points) What is the critical value of the relevant test‐statistic?    2 df = (r − 1)(c − 1), df = (2 − 1)(2 − 1), df = 1, for α = 0.05 χ critical = 3.84       c. (3 points) What are the expected values?    ff fe = c r   n     Expected  Opinion    Favor  Oppose  City  60*50/150 = 20  Suburb        90*50/150 = 30  90*100/150 = 60  60*100/150 = 40      d. (5 points) Do the data indicate a statistically significant difference in the preference  of those people living in the city and those people living in the suburbs? Test using  α = 0.05  and explain what this means for the original research question.       χ 2 observed ⎛ ( f o − f e )2 ⎞ = ∑⎜ ⎟ f ⎝ ⎠ e ⎛ ( 35 − 30 )2 ⎞ ⎛ (15 − 20 )2 ⎞ ⎛ (55 − 60 )2 ⎞ ⎛ ( 45 − 40 )2 ⎞ 2 χ observed = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ 30 20 60 40 ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎛ 25 ⎞ 2 χ observed = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 30 ⎠ ⎝ 20 ⎠ ⎝ 60 ⎠ ⎝ 40 ⎠   2 χ observed = 0.833 + 1.25 + 0.4166 + 0.625 2 χ observed = 3.125       2 2 χ observed = 3.125, χ critical = 3.84 2 2 χ observed < χ critical , Fail to reject H 0   The data do not provide evidence that opinions are related to the region in which  individuals live.     ...
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## This note was uploaded on 10/16/2011 for the course PSYCH 60 taught by Professor Parris during the Spring '11 term at UCSD.

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