Question

In: Statistics and Probability

. A nutritionist found that in a sample of 80 families, 22% said they ate apples at...

. A nutritionist found that in a sample of 80 families, 22% said they ate apples at least once a week. Find

the 90% confidence interval of the true proportion of families who eat apples at least once a week.

5. A survey of 120 female freshmen showed that 18 did not wish to work after marriage. Find the 95%

confidence interval of the true proportion of females who do not wish to work after marriage.

Expert Solution

4)

Given,

= 0.22 , n = 80

90% confidence interval for p is

- Z/2 * Sqrt ((1-) / n) < p < + Z/2 * Sqrt ((1-) / n)

0.22 - 1.645 * sqrt( 0.22 * 0.78 / 80) < p < 0.22 + 1.645 * sqrt( 0.22 * 0.78 / 80)

0.144 < p < 0.296

90% CI is ( 0.144 , 0.296 )

5)

sample proportion = 18 / 120 = 0.15

95% confidence interval for p is

- Z/2 * Sqrt ((1-) / n) < p < + Z/2 * Sqrt ((1-) / n)

0.15 - 1.96 * sqrt ( 0.15 * 0.85 / 120) < p < 0.15 + 1.96 * sqrt ( 0.15 * 0.85 / 120)

0.086 < p < 0.214

95% CI is ( 0.086 , 0.214 )

orchestra answered 2 years ago

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