There is variation in the amount of water put in 24-ounce bottles when filled on the assembly line. Amounts are skewed left with mean,
𝜇
=
24.1
oz, and standard deviation
𝜎
=
0.4
oz. If repeated random samples of size 36 bottles are pulled off the assembly line, what will the standard error of the sampling distribution of
𝑋
¯
be? Round your answer to the nearest hundredths place.

Question

There is variation in the amount of water put in 24-ounce bottles when filled on the assembly line. Amounts are skewed left with mean,  
𝜇
=
24.1
 oz, and standard deviation 
𝜎
=
0.4
 oz. If repeated random samples of size 36 bottles are pulled off the assembly line, what will the standard error of the sampling distribution of 
𝑋
¯
 be? Round your answer to the nearest hundredths place.

BlackTom AI · Accepted Answer

The standard error of the sample mean is σ/√n, so we compute it directly using σ = 0.4 oz and n = 36.

1. Relevant knowledge points and concepts
- Sampling distribution of the sample mean: the mean of sample means equals the population mean μ, and the standard deviation of the sampling distribution (the standard error) is σ/√n.
- Central Limit Theorem: for sufficiently large n (commonly n ≥ 30), the sampling distribution of X̄ is approximately normal even if the population distribution is skewed. This justifies using the standard error formula here despite left skewness.
- Units: the standard error will be in ounces.

2. Complete reasoning process (step by step)
- Start with the formula for the standard error of the mean: SE = σ / √n.
- Substitute the given values: σ = 0.4 oz, n = 36.
- Compute √n and divide σ by that value to get SE.

3. Calculations with intermediate steps
- √n = √36 = 6.
- SE = 0.4 / 6.
- 0.4 ÷ 6 = 0.066666... oz.
- Round to the nearest hundredths place: 0.066666... → 0.07 oz.

4. Why the correct answer is correct
- The formula SE = σ/√n is derived from properties of variance for independent observations: Var(X̄) = Var((1/n)ΣXi) = (1/n^2) n σ^2 = σ^2/n, so SD(X̄) = σ/√n. Using σ = 0.4 and n = 36 yields 0.4/6 = 0.066666..., which rounds to 0.07. The sample size of 36 is large enough that skewness does not invalidate using this formula for the sampling distribution.

5. Other possible mistakes (why common alternatives would be incorrect)
- Dividing by n instead of √n (0.4/36 ≈ 0.0111) would be incorrect because variance scales with 1/n not 1/n^2.
- Forgetting to take the square root of n or using the population mean instead of σ would yield wrong units or values.
- Rounding prematurely to 0.06 would be incorrect when rounding to the nearest hundredths; correct rounding gives 0.07.

6. Conclusion and final answer confirmation
- The standard error of the sampling distribution of X̄ is 0.066666... oz, which rounds to 0.07 oz. This matches the provided correct answer: 0.07.

类似问题

What is a sampling distribution?

If you increase the sample size from 25 to 100 while keeping the population standard deviation constant, what happens to the standard deviation of the sampling distribution of the mean?

Which of the following statements is correct about a parameter or a statistic associated with repeated random samples of the same size from the same population?

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