Question

1. Test the claim about the population mean, ?, at the level of significance, ?. Assume the population is normally distributed.

Claim: ? ≤ 47, ? = 0.01, ? = 4.3 Sample statistics: ?̅ = 48.8, ? = 40

A. Fail to reject ?0. There is enough evidence at the 1% significance level to support claim.

B. Not enough information to decide.

C. Reject ?0. There is enough evidence at the 1% significance level to reject the claim.

2.

Use the following information to answer questions 28, 29, and 30. The heights in inches of 10 US adult males are listed below.

70 72 71 70 69 73 69 68 70 71

a) Determine the range.

b) Determine the standard deviation.

c) Determine the variance.

3. The weights in pounds of 30 preschool children are listed below. Find the five number summary of the data set.

25 25 26 26.5 27 27 27.5 28 28 28.5

29 29 30 30 30.5 31 31 32 32.5 32.5

33 33 34 34.5 35 35 37 37 38 38

4. A manufacturer receives an order for light bulbs. The order requires that the bulbs have a mean life span of 850hours. The manufacturer selects a random sample of 25 light bulbs and finds they have a mean life span of 845 hours with a standard deviation of 15 hours. Assume the data are normally distributed. Using a 95% confidence level, test to determine if the manufacturer is making acceptable light bulbs and include an explanation of your decision.

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.

Answer 1

Solution

Q1

Given population is N(µ, σ²)

Claim: µ ≤ 47

Hypotheses:

Null H₀: µ = µ₀ = 47   Vs Alternative H_A: µ > 47

Test statistic:

Z = (√n)(Xbar - µ₀)/σ, = 2.647

where

n = sample size;

Xbar = sample average;

σ = known population standard deviation.

Summary of Excel Calculations is given below:

Given, n	40
µ0	47
σ	4.3
Xbar	48.8
Zcal	2.647488
Given α	0.1
Zcrit	1.281552
p-value	0.004055

Distribution, Level of Significance, α, Critical Value and p-value

Under H₀, Z ~ N(0, 1)

Critical value = upper (α/2)% point of N(0, 1).

p-value = P(Z > | Zcal |)

Using Excel Functions: Statistical NORMSINV and NORMSDIST, Zcrit and p-value are found to be as shown in the above table.

Decision:

Since Zcal > Zcrit, or equivalently, since p-value < α. H₀ is rejected.

Conclusion:

Reject ?0. There is enough evidence at the 1% significance level to reject the claim.

Option C Answer

DONE

Q2

Max	73
Min	68
Range	5
Standard Deviation	1.418
Variance	2.010724

Answer

Done