Question

10. State which is the most appropriate: z or t or neither. Give the critical value when you can.

            95% confidence; ; ; population has an outlier

11. We want to know if the percentage of last-born children who are the most assertive has changed from the 43% estimate from a study in 2017. How many families should be selected to produce an estimate that is within 1.2% of the population percentage with 98% confidence?

Answer 1

solution

10.

when sample size less than 30 and sample standard deviation is given we use t

when sample size greater than 30 and populationstandard deviation is given we use z

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

11.

Solution :

Given that,

= 0.43

1 - = 1 - 0.43 = 0.57

margin of error = E = 1.2% = 0.012

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = 2.33 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.33 / 0.012)2 * 0.43 * 0.57

=9240.44

Sample size = 9241