An overly involved "neighborhood watch" group has been investigating the length of lawns. In years past, lawn lengths are normally distributed and lawns had been mowed to a mean length of 3.5 inches. A recent random sample of 16 lawn lengths is given below. Conduct an appropriate test, at a 5% significance level, to determine if the overall mean lawn length is now higher than 3.5 inches. Round all answers to 4 decimal places. Lawn Length (in.) 3.14 3.64 3.49 3.04 4.09 4.54 3.86 4.27 5.01 3.9 4.77 3.54 3.97 2.51 4.08 3.39 Checksum: 61.24 a. Consider the parameter of interest and choose the correct alternate hypothesis. μ≠3.5μ≠3.5 μ<3.5μ<3.5 μ=3.5μ=3.5 μ>3.5μ>3.5 b. Are the necessary conditions present to carry out this inference procedure? Explain in context. Our sample was not larger than 30 in size, but it is stated that the sample was randomly gathered from a normally distributed population. No. The sample is not large enough and a histogram of the data shows the population is uniform. Carry out the procedure (crunch the numbers using technology): c. ¯x≈x¯≈ d. Standard deviation: e. Test statistic: t≈t≈ f. P-value: g. Decision: Reject the claim. Accept the null hypothesis. Reject the null hypothesis. Fail to reject the null hypothesis. h. Write a conclusion in context. Because our p-value is less than our level of significance (alpha=0.05), we have sufficient evidence to reject the null hypothesis. The evidence seems to indicate that the overall mean lawn lengthis higher than 3.5 inches. Because our p-value is not less than our level of significance (alpha=0.05), we do not have sufficient evidence to reject the null hypothesis. We cannot conclude that the overall mean lawn length is higher than 3.5 inches.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

An overly involved "neighborhood watch" group has been investigating the length of lawns. In years past, lawn lengths are normally distributed and lawns had been mowed to a mean length of 3.5 inches. A recent random sample of 16 lawn lengths is given below. Conduct an appropriate test, at a 5% significance level, to determine if the overall mean lawn length is now higher than 3.5 inches. Round all answers to 4 decimal places.

Lawn Length (in.)
3.14
3.64
3.49
3.04
4.09
4.54
3.86
4.27
5.01
3.9
4.77
3.54
3.97
2.51
4.08
3.39

Checksum: 61.24

a. Consider the parameter of interest and choose the correct alternate hypothesis.

μ≠3.5μ≠3.5
μ<3.5μ<3.5
μ=3.5μ=3.5
μ>3.5μ>3.5

b. Are the necessary conditions present to carry out this inference procedure? Explain in context.

Our sample was not larger than 30 in size, but it is stated that the sample was randomly gathered from a normally distributed population.
No. The sample is not large enough and a histogram of the data shows the population is uniform.

Carry out the procedure (crunch the numbers using technology):

c. ¯x≈x¯≈

d. Standard deviation:

e. Test statistic: t≈t≈

f. P-value:

g. Decision:

Reject the claim.
Accept the null hypothesis.
Reject the null hypothesis.
Fail to reject the null hypothesis.

h. Write a conclusion in context.

Because our p-value is less than our level of significance (alpha=0.05), we have sufficient evidence to reject the null hypothesis. The evidence seems to indicate that the overall mean lawn lengthis higher than 3.5 inches.
Because our p-value is not less than our level of significance (alpha=0.05), we do not have sufficient evidence to reject the null hypothesis. We cannot conclude that the overall mean lawn length is higher than 3.5 inches.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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