A packaging company conducts a survey for the two new package designs for lunch boxes in a supermarket. The results are as follows: Design1 158.5 138.4 168.1 149.4 145.8 168.7 154.4 162.9 Design2 150.3 155.4 151.6 158.8 151.4 150.8 161.4 157.6 156.8 147.6 a. What is the range of the true (or population) mean of the data in each design for 80% confidence level

Design 1: 151.19≤x≤160.37

Design 2: 152.46≤x≤155.88

Step-by-step explanation:

Confidence interval formula is expressed as;

CI = xbar ± Z×σ/√n

xbar is the mean of the sample

σ is the standard deviation

n is the sample size

z is the z score at 80% confidence level

For Design 1:

158.5 138.4 168.1 149.4 145.8 168.7 154.4 162.9

xbar = Sum of the samples/sample size

Sum of samples = 158.5+ 138.4+ 168.1+149.4+ 145.8+ 168.7+ 154.4+ 162.9

Sum of samples = 1246.2

Sample size = 8

xbar = 1246.2/8

xbar = 155.78

Standard deviation

σ = √\sum(x-xbar)²/N

σ = (158.5 - 155.775)² + ... + (162.9 - 155.775)²/8

= 821.075/8

= 102.634375

= √102.634375

= 10.13

σ = 10.13

CI = 155.78±(1.282×10.13/√8)

CI = 155.78±(1.282×3.5815)

CI = 155.78±(4.5915)

CI = {155.78-4.5915, 155.78+4.5915}

CI = {151.19, 160.37}

The range for the true mean is 151.19≤x≤160.37

For Design 2:

150.3 155.4 151.6 158.8 151.4 150.8 161.4 157.6 156.8 147.6

xbar = Sum of the samples/sample size

Sum of samples = 150.3+ 155.4+ 151.6+158.8 +151.4+ 150.8+ 161.4 +157.6+ 156.8 +147.6

Sum of samples = 1541.7

Sample size = 10

xbar = 1541.7/10

xbar = 154.17

Standard deviation

σ = (150.3 - 154.17)²+ ... + (147.6 - 154.17)²/10

σ = 177.681/10

σ = 17.7681

σ = √17.7681

σ = 4.22

CI = 154.17±(1.282×4.22/√10)

CI = 154.17±(1.282×1.3345)

CI = 154.17±(1.7108)

CI = {154.17-1.7108, 154.17+1.7108}

CI = {152.46, 155.88}

The range for the true mean is 152.46≤x≤155.88

i believe your answer is 99.7% of 7-year-old children are between 46.6 inches and 49.5 inches tall.