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Yuki888 [10]
4 years ago
11

the sample mean foot length of a simple random sample of 25 third-graders is 22.5 cm. The standard error of the mean is 0.8 cm.

Which one of the following is a correct interpretation for the standard error of the mean?
Mathematics
1 answer:
GenaCL600 [577]4 years ago
8 0

Question:

The sample mean foot length of a simple random sample of 25 third-graders is 22.5 cm. The standard error of the mean is 0.8 cm. Which one of the following is a correct interpretation for the standard error of the mean?

The typical distance between each individual foot length in the sample and the sample mean foot length is approximately 0.8 cm.

The typical distance between one sample mean foot length and another sample mean foot length is 0.8 cm.

The typical distance between each individual foot length in the population and the true mean foot length is approximately 0.8 cm.

The typical distance between means of samples of size 25 and the population mean foot length is approximately 0.8 cm

The typical distance between each individual foot length in the sample and the true mean foot length is approximately 0.8 cm.

Answer:

The correct option is:

The typical distance between means of samples of size 25 and the population mean foot length is approximately 0.8 cm

Step-by-step explanation:

The standard error of the mean is a measure of the extent of variation of the mean in a sample of data from that of the value of the mean within the general population. It is a statistical measure of the probability that the size of a particular sample will approach that of the population mean, with reference to the principle of the central limit theorem.

Therefore, the interpretation of the standard error is that the variation between mean in a sample of 25 and that of the population is about 0.8 cm.

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Answer:

Step-by-step explanation:

Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.7875

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Summation(x - mean)² = (29.7 - 29.7875)^2 + (29.4 - 29.7875)^2 + (31.7 - 29.7875)^2 + (29.0 - 29.7875)^2 + (29.1 - 29.7875)^2 + (30.5 - 29.7875)^2 + (29.1 - 29.7875)^2 + (29.8 - 29.7875)^2 = 5.88875

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a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0:µ ≤ 30000

For the alternative hypothesis,

H1:µ > 30000

This is a right tailed test.

b) The decision rule is to reject the null hypothesis if the significance level of 0.05 is greater than the probability value. If it is otherwise, we would fail to reject the null hypothesis.

c) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

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Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29787.5

µ = population mean = 30000

s = samples standard deviation = 880

t = (29787.5 - 30000)/(880/√8) = - 0.68

We would determine the p value using the t test calculator. It becomes

p = 0.26

d) Since alpha, 0.1 < the p value, 0.26, then we would fail to reject the null hypothesis.

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