A consumer advocacy goup is ivestignting sugar contcnt in a particular brand of

Question

A consumer advocacy goup is ivestignting sugar contcnt in a particular brand of cranberry sauee. A random sample of 50 servings was selected and sugar content was measured. 6 (12p) The mean amount of sugar per serving was found to be 25.4 g and the sample standard deviation was found to be 1.8 g. I for the Mean amount of sugar per serving b. The sample result of & 25.4 g can be used as a point estimate for the amount of sugar per serving. At the 95% Conf fidence Level, what is the Margin of Error associated with this estimate?

Explanation / Answer

a, HERE,

n = 50

mean amount of sugar per serving = 24.4 g

standard deviation = 1.8

95% CI = mean ± margin of error

margin of error = critical value X standard error

Here we will use the t distribution for critical value as the population standard deviation is now know

critical value for 50 sample size, degree of freedom 49 (50-1) for 95% CI = 2.009

standard error = sd/n = 1.8/ 50 = 1.8/ 7.071 = 0.254

therefore, margin of error = critical value X standard error = 2.009 X 0.254 = 0.51

95% CI = mean ± margin of error

Lower limit of 95% CI = 25.4 - 0.51 = 24.89

Upper limit of 95% CI = 25.4 + 0.51 = 25.91

b. from the above example, margin of error = 0.51

Analysis from STATA softwae is given below

. cii 50 25.4 1.8

Variable | Obs Mean Std. Err. [95% Conf. Interval]
-------------+---------------------------------------------------------------
| 50 25.4 .2545584 24.88845 25.91155

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