Question

In: Advanced Math

5. A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large...

5. A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large of a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 4%.

6.

A local group claims that the police issue at least 60 speeding tickets a day in their area. To prove their point, they randomly select one month. Their research yields the number of tickets issued for each day. The data are listed below. Assume the population standard deviation is 12.2 tickets. At ? = 0.01, test the group’s claim. Make sure to state your conclusion regarding the claim with your reasoning.

70 48 41 68 69 55 70 57 60 83 32 60 72 58 88 48

59 60 56 65 66 60 68 42 57 59 49 70 75 63 44

7. A local politician, running for reelection, claims that the mean prison time for car thieves is less than the required 4 years. A sample of 80 convicted car thieves was randomly selected, and the mean length of prison time was found to be 3.5 years. Assume the population standard deviation is 1.25 years. At ? = 0.05, test the politician’s claim. Make sure to state your conclusion regarding the claim with your reasoning.

Solutions

Expert Solution

5) Given confidence level = 95%

alpha = 5%

margin of error E = 0.04

sample size calculation

p = 0.5

Z0.025 = 1.96

n = = 600.25 = 600

6)

One-Sample Z: Data

Descriptive Statistics

N Mean StDev
31 60.39 12.19

μ: mean of Data
Known standard deviation = 12.2

Test

Null hypothesis H₀: μ = 60
Alternative hypothesis H₁: μ < 60

? = 0.01

test statistic Z = = 0.18

Z-Value P-Value
0.18 0.570

critical value = -Z0.01 = -2.33

conclusion: Since the test statistic is greater than critical value we fail reject null hypothesis and there is a no significant evidence to conclude that the number of tickets issued is less than 60tickets per day.

7) Given mean = 3.5

= 1.25

Hypothesis

H0 : = 4

H1 : < 4

? = 0.05

test statistic Z = = -3.577.

critical value = -Z0.05 = -1.645

conclusion: Since the test statistic is lesser than critical value we reject null hypothesis and there is a significant evidence to conclude that the mean prison time for car thieves is less than the required 4 years.


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