Question

180 students were asked to randomly pick one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The number 7 was picked by 61 students.

a) For the sample, calculate the proportion of students who picked 7. (Round the answer to three decimal places.)

(b) Calculate the standard error for this sample proportion. (Round the answer to three decimal places.)

(c) Calculate a 90% confidence interval for the population proportion. (Round the answer to three decimal places.) to

(d) Calculate a 95% confidence interval for the population proportion. (Round the answer to three decimal places.) to

(e) Calculate a 98% confidence interval for the population proportion. (Round the answer to three decimal places.) to

(f) What do the results of parts (c)-(e) indicate about the effect of confidence level on the width of a confidence interval? (Select all that apply.)

As the confidence level is increased, the width of the interval decreases.

As the confidence level is decreased, the width of the interval decreases.

As the confidence level is decreased, the width of the interval increases.

As the confidence level is increased, the width of the interval increases.

(g) On the basis of these confidence intervals, do you think that students choose numbers "randomly"?

Yes

No

Answer 1

Solution-A:

sample proprtion=those who picked 7/total

=succeses/total

=61/180

=0.3388889

=0.339

sample proportion=0.339

(b) Calculate the standard error for this sample proportion. (Round the answer to three decimal places.)

standrard error=sqrt(p^*(1-p^)/n)

=sqrt(0.3388889*(1-0.3388889)/180 )

=0.03528009

=0.035

ANSWER;

standard error=0.035

(c) Calculate a 90% confidence interval for the population proportion. (Round the answer to three decimal places.) to

z crit for 90%=1.645

90% confidence interval for the population proportion is

p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

0.3388889-1.645*sqrt(0.3388889*(1-0.3388889)/180 ,0.3388889+1.645*sqrt(0.3388889*(1-0.3388889)/180 )

0.2808532,0.3969246

0.281,0.397

0.281,0.397

(d) Calculate a 95% confidence interval for the population proportion. (Round the answer to three decimal places.) to

z crit for 95%=1.96

95% confidence interval for the population proportion is

p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

0.3388889-1.96*sqrt(0.3388889*(1-0.3388889)/180 ,0.3388889+1.96*sqrt(0.3388889*(1-0.3388889)/180 )

0.2697399, 0.4080379

0.270,0.408

0.270,0.408