Sample Mean and Standard Deviation

(a) (iii) Set of income responses from all female in the US 

(b) (i) Set of income responses from 3500 females in the US

Justification for (a) and (b):

(a) (iii) is a parameter since it represents the median of the population and not of the sample. Parameter is representative of characteristic of the population.(b) (i) is a sample since it captures the observations from certain selected females in the US.

1. a) (iii) Ordinal

(b) (iii) Systematic

Justification for (a) and (b):

(a) STUDENT ID numbers are considered as ordinal data since this is essentially categorical data represented in numerical form and can be arranged in an orderly manner.

(b) This is systematic sampling since every 100^th product is chosen from the sample line. In systematic sampling, every nth item is chosen where n is essentially a random number which once determined does not alter.

(b) Percentage of midterm exam scores at least 80

Work for (a) and (b):

(a) Range = 53

(b) Median =83.5

(c) Mode = 84

Work for (a), (b) and (c):

Ascending order of the data

) Sample mean = 75.70

(b) Sample standard deviation = 12.89

(c) Excel function has been used to determine the sample mean and standard deviation. The AVERAGE () function and STDEV.S() function would provide the mean and standard deviation of the sample.

Answer: 

(a) Probability of drawing a diamond or clover = 1/2

(b) Probability that the card is not a spade = 3/4

Work for (a) and (b):

Probability of drawing a diamond or clover

Probability that the card is not a spade  

(a) Probability that first ball is white and second ball is also white (when ball selection is with replacement) = 0.36

(b)  Probability that first ball is white and second ball is also white (when ball selection is without replacement) = 0.33

Work for (a) and (b): 

1. a) Probability that the first ball is white = (6/10)

Now since replacement is done, hence again there are 6 white balls and 4 red balls.

Probability that the second ball is white = (6/10)

Hence, requisite probability = (6/10)*(6/10) = 0.36

1. b) Probability that the first ball is white = (6/10)

Now since replacement is not done, hence there are 5 white balls and 4 red balls.

Probability that the second ball is white = (5/9)

Hence, requisite probability = (6/10)*(5/9)p= 0.33

Answer:    

(a) Yes, the order does matter in scheduling. This is because the assistant may select any of the 5 stores among the 20 stores. It means it would become a case of combination. However, while scheduling the visits, the order on which the stores are going to be visited would be different.

Drawing Cards from a Deck

(b) Permutation would be used in order to determine the different schedule that assistance may have.

(c) Number of different schedules can assistant recommend =1,860,480

Work for (c): 

Number of different schedules can assistant recommend

(a) No

(b) Combination

(c) 56

Work for (c):

Total number of books (n) = 8

Selection of books (r) = 3

Different ways of selecting r books of combination = nCr

Different ways of selecting 3 books of combination = 56

1. b) mean =5         ,  standard deviation =0.87             .                          

Work for (a) and (b):

Fair coin is tossed 3 times then outcomes

{HHH, HHT, HTH, THH, THT, TTH, HTT, TTT}

When trials are independent then, P (H) =P(T) =1/2, P(HHH)=1/8, P(HHT)=1/8, P(HTH)=1/8, P(THH)=1/8, P(THT)=1/8, P(TTH)=1/8, P(HTT)=1/8, P(TTT)=1/8

Let X is random variable denoting number of heads then,

X(0) = TTT,   X(1)= THH, TTH, THT,  X(2) =THH,HHT, HTH  ,  X(3) =HHH

Mean  

Standard deviation

nswer:  

(a) P(8.5<x<12.5) = 0.6678

(b) 90^th percentile = 12.56

(c) An online tool has been taken into account to find the probability. Further, excel function has used to find the percentile of pecan tree height distribution (90^th percentile =NORMSINV(0.9))

Work for (a) and (b): 

1. a)95% confidence interval = [0.5068    0.5557]

(b) There is a 95% confidence that proportion of adult respondents who live in a household without landline would fall between the range of 0.5068 and 0.5557.

Work for (a): 

Excel supporting tool KADD has been taken into consideration to determine the 95% confidence interval for population proportion. The output is shown below.

(a) One sample z test for population proportion.

• Null hypothesis H₀: p <= 0.50

Alternative hypothesis H1: p>0.50

• Z stat = 2.50
• The p value = 0.0124

(e)  It can be said that p value (0.0124) is lower than level of significance (5%) and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis.

(f )Hence, sufficient evidence is present to conclude that proportion of adults who live in a households without  landline phones is higher than 50%.

Work for (c) and (d):

Online calculator has been used to find the z and p value.

• The t test for two dependent samples (matched pairs) because the data 1 hours later and 24 hours later are dependent and hence, matched pair would be used to check the validity of the claim.

(e) the p value (one tailed hypothesis test) =0.017

• It can be said that p value (0.017) is lower than level of significance (5%) and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis.(g)Hence, sufficient evidence is present to conclude that mean number of words recalled after 1 hours would exceed the mean recall after 24 hours because alternative hypothesis has been accepted.

Work for (d) and (e):

Excel output for performing the t test for paired two sample

• Chi-square test (color distribution and color are the two variable and hence, the cross-tabulation table would be used to test the hypothesis which involves use of chi-square test).
• Null hypothesis H₀: True proportion in all the five categories is same to their expected value.

Alternative hypothesis H1: At least one proportion is different from their expected value.

• Test statistic
• The p value = 0.00
• It can be said that p value is lower than level of significance and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis.
• There is sufficient evidence is present to support the claim that published color distribution is correct because null hypothesis has rejected.

• Least square regression line

(b) Final exam score for 80 average quiz score =79.81

(c) Final exam score for 40 average quiz score =45.21

(d) Final exam score for 80 average quiz score which is 79.81 is more closer to true final exam score.

Work for (a), (b), and (c):

Scatter plot has been made between x and y through excel and also fitted the regression line.

 t test for two independent sample

(b) ANOVA

Justification for (a) and (b): 

The t test for two independent sample would be used when the population variance is not given. The t test for dependent samples would be used when population variance is known. Further, Chi square is generally used when there is two-way table is provided for the given two variables only. ANOVA is used when there is at least three random gatherings.