Question

It is recommended that the daily intake of sodium be 2600 mg per day. From a previous study on a particular ethnic group, the prior distribution of sodium intake is believed to be normal with mean 3000 mg and standard deviation 300 mg. If a recent survey resulted in a mean of 3500 mg and standard deviation of 350 mg. Obtain a 99% Credible Interval for the mean intake of sodium for this ethnic group. Let n=1

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You don't have to solve this because I know how to solve for a simple confidence interval. My question is, which distribution do we use for this when the sample size is 1? My first instinct is to use t, but what would be the degrees of freedom?

Answer 1

Answer to the question)
This is the scenario of one sample mean confidence interval

It can either be Z confidence interval or T confidence interval

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Now when the population standard deviation (σ) value is provided, we always use the Z confidence interval formula

If the value of population standard deviation ( σ) is not provided, only then we switch to the T confidence interval formula

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The formulae for both the interval are exactly same, just the t critical value is replaced by Z critical value

in case of Z critical value you need to depend on the degree of freedom value

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Solution:

Formula of Z confidence interval is:

(x bar - z * σ/ sqrt( n) ) . ( x bar + z * σ/ sqrt( n) )

where:

x bar = sample mean = 3500 mg

σ = 300 mg [ population standard deviation is provided in the question]

n = sample size = 1

Value of Z critical value for 99% confidence level is:

(b) Confidence Interval Critical Values z Critical Value z 1.04 1.15 1.28 1.44 1.645 1.96 2.33 2.58 Level of Confidence c 0.70, or 70% 0.75, or 75% 0.80, or 80% 0.85, or 85% 0.90, or 90% 0.95, or 95% 0.98, or 980 0.99, 99%

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Thus we get Z 2.58

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On plugging these values we get:

(3500 -2.58 * 300/sqrt( 1) , 3500 +2.58 *300/sqrt(1) )

(2726 , 4274)