Question

In: Statistics and Probability

The following data represent crime rates per 1000 population for a random sample of 46 Denver...

The following data represent crime rates per 1000 population for a random sample of 46 Denver neighborhoods.† 63.2 36.3 26.2 53.2 65.3 32.0 65.0 66.3 68.9 35.2 25.1 32.5 54.0 42.4 77.5 123.2 66.3 92.7 56.9 77.1 27.5 69.2 73.8 71.5 58.5 67.2 78.6 33.2 74.9 45.1 132.1 104.7 63.2 59.6 75.7 39.2 69.9 87.5 56.0 154.2 85.5 77.5 84.7 24.2 37.5 41.1 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to one decimal place.) x = crimes per 1000 people s = crimes per 1000 people (b) Let us say the preceding data are representative of the population crime rates in Denver neighborhoods. Compute an 80% confidence interval for μ, the population mean crime rate for all Denver neighborhoods. (Round your answers to one decimal place.) lower limit crimes per 1000 people upper limit crimes per 1000 people (c) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 59 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate is below the average population crime rate. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. No. The confidence interval indicates that this crime rate is below the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (d) Another neighborhood has a crime rate of 74 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (e) Compute a 95% confidence interval for μ, the population mean crime rate for all Denver neighborhoods. (Round your answers to one decimal place.) lower limit crimes per 1000 people upper limit crimes per 1000 people (f) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 59 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate is below the average population crime rate. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. No. The confidence interval indicates that this crime rate is below the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (g) Another neighborhood has a crime rate of 74 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. Yes. The confidence interval indicates that this crime rate does not differ from the average population crime rate. Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate is higher than the average population crime rate. No. The confidence interval indicates that this crime rate does not differ from the average population crime rate. (h) In previous problems, we assumed the x distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: Use the central limit theorem. Yes. According to the central limit theorem, when n ≥ 30, the x distribution is approximately normal. Yes. According to the central limit theorem, when n ≤ 30, the x distribution is approximately normal. No. According to the central limit theorem, when n ≥ 30, the x distribution is approximately normal. No. According to the central limit theorem, when n ≤ 30, the x distribution is approximately normal.

Solutions

Expert Solution

(a) μ = 64.2 and σ = 27.9

(b)

n = 46     

x-bar = 64.2     

s = 27.9     

% = 80     

Standard Error, SE = s/√n =    27.9/√46 = 4.113630577

Degrees of freedom = n - 1 =   46 -1 = 45   

t- score = 1.300649332     

Width of the confidence interval = t * SE =     1.30064933223503 * 4.11363057672163 = 5.350390863

Lower Limit of the confidence interval = x-bar - width =      64.2 - 5.35039086267459 = 58.84960914

Upper Limit of the confidence interval = x-bar + width =      64.2 + 5.35039086267459 = 69.55039086

The confidence interval is [58.8, 69.6]

(c) No. The confidence interval indicates that this crime rate does not differ from the average population crime rate

(d) Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate

(e) No. According to the central limit theorem, when n ≥ 30, the x distribution is approximately normal

n = 46     

x-bar = 64.2     

s = 27.9     

% = 95     

Standard Error, SE = s/√n =    27.9/√46 = 4.113630577

Degrees of freedom = n - 1 =   46 -1 = 45   

t- score = 2.014103359     

Width of the confidence interval = t * SE =     2.01410335926697 * 4.11363057672163 = 8.285277163

Lower Limit of the confidence interval = x-bar - width =      64.2 - 8.28527716335835 = 55.91472284

Upper Limit of the confidence interval = x-bar + width =      64.2 + 8.28527716335835 = 72.48527716

The confidence interval is [55.9, 72.3]

(f) No. The confidence interval indicates that this crime rate does not differ from the average population crime rate

(g) Yes. The confidence interval indicates that this crime rate is higher than the average population crime rate

(h)


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