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pishuonlain [190]
3 years ago
9

The number of hours that a nine month old baby sleeps at night are normally distributed with a population standard deviation of

1.5 hours and an unknown population mean. A random sample of 22 nine month old babies is taken and results in a sample mean of 12 hours. Find the margin of error for a confidence interval for the population mean with a 90% confidence level.
Mathematics
1 answer:
natita [175]3 years ago
3 0

Answer:

The margin of error is of 0.7123 hours.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

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

In which \sigma is the standard deviation of the population and n is the size of the sample.

Find the margin of error for a confidence interval for the population mean with a 90% confidence level.

We have that \sigma = 1.5, n = 12. So

M = 1.645*\frac{1.5}{\sqrt{12}} = 0.7123

The margin of error is of 0.7123 hours.

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<h2><u>Answer with explanation:</u></h2>

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu

, where n= sample size

\overline{x} = Sample mean

s= sample size

t* = Critical value.

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Degree of freedom : df=n-1=24

\overline{x}= \$93.36

s=\ $19.95

Significance level for 98% confidence interval : \alpha=1-0.98=0.02

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu

=93.36-9.943878< \mu

=93.36-9.943878< \mu

=83.416122< \mu

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