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Rudik [331]
3 years ago
6

Suppose you have a sample of size 42 with a mean of 30 and a population standard deviation of 7.4. What is the maximal margin of

error associated with a 95% confidence interval for the true population mean? As in the reading, in your calculations, use z = 2. Give your answer as a decimal, to two places
Mathematics
1 answer:
9966 [12]3 years ago
4 0

Answer:

The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.

Step-by-step explanation:

We have given,

The sample size n=42

The sample mean \bar{x}=30

The population standard deviation \sigma=7.4

Let \alpha be the level of significance = 0.05

Using the z-distribution table,

The critical value at 5% level of significance and two tailed z-distribution is

\pm z_{\frac{0.05}{2}}=\pm 1.96

The value of margin of error is

ME=z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})

ME=1.96(\frac{7.4}{\sqrt{42}})

ME=1.96(1.1418)

ME=2.238

The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.

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