Question

A survey commissioned by the Southern Cross Healthcare Group reported that 15 % of New Zealanders consume five or more servings of soft drinks per week. The data were obtained by an online survey of 2090 randomly selected New Zealanders over 15 years of age (a) What number of survey respondents reported that they consume five or more servings of soft drinks per week? You will need to round your answer. XX = (b) Find a 95% confidence interval ( ±±0.001) for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. 95% confidence interval is from to (c) Convert the estimate of your confidence interval to percents ( ±±0.1) % to %

Answer 1

Solution :

Given that,

n = 2090

= 15% = 0.15

1 - = 1 - 0.15 = 0.85

a) x = * n = 0.15 * 2090 = 314

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025  = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.15 * 0.85) / 2090)

= 0.015

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.15 - 0.015 < p < 0.15 + 0.015

( 0.135 < p < 0.165 )

c) ( 13.5% < p < 16.5% )