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IrinaK [193]
3 years ago
6

To use the two sample t procedure to perform a significance test on the difference of two means, we assume: a) The populations'

standard deviation are known. b) The samples from each population are independent. c) The sample sizes are large. d) The distributions are exactly normal in each population.
Mathematics
1 answer:
Anika [276]3 years ago
7 0

Answer:

The option which is an exception is;

c) The sample sizes are large

Step-by-step explanation:

Here the requirement to perform a two sample t procedure are as follows;

1. The two samples have equal variance, hence the variance should be known

2. The samples for which the two sample t procedure is performed must be independent

3. The data are in alignment with the normal distribution probability

4. The samples are random samples taken from the different respective population

Since the sample size is known, we do not assume that the sample size are large.

Therefore options a), b) and d) are all correct except option c).

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<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
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For this problem, the parameters are:

z = 1.96, \sigma = 40, M = 6

Hence we solve for n to find the needed sample size.

M = z\frac{\sigma}{\sqrt{n}}

6 = 1.96\frac{40}{\sqrt{n}}

6\sqrt{n} = 40 \times 1.96

\sqrt{n} = \frac{40 \times 1.96}{6}

(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2

n = 170.7.

Rounding up, a sample of 171 should be selected.

More can be learned about the z-distribution at brainly.com/question/25890103

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