Question

STT242-sampleProbleA OMOLAYO/Desktop/STAT244-Sampl eProblems-Q Favorites Bar folder. Or import from another browser. Import rcise 8. the vast majority is dependent upon it. Suppose that in a survey of 500 female and 500 e students in Generation Next, 345 of the females and 365 of the males reported that y decided to attend university in order to make more money KY (a) Construct a 90% confidence interval for the difference in the proportions of fmale and male students who decided to attend university in order to make money. (Be sure to (b) Based on the confidence interval in part (a), can you conclude that there is a difference in the proportions of fomuale anul male students who deidel to attend university in state the conditions required in order for this confidence interval to be valid.) Et

Answer 1

$=\frac{345+365}{500+500}=0.71$

Q= 1 - P = 0.29

$SE=\sqrt{PQ(1/n1+1/n2)}$

$=\sqrt{0.71\times 0.29(1/500+1/500)}=0.0287$

$\alpha$ = 0.10

From Table, critical value of Z = $\pm$ 1.645

p1 = 345/500 = 0.69

p2 = 365/500 = 0.73

Confidence interval:

(0.69 - 0.73) $\pm$ (1.645 X 0.0287)

= - 0.04 $\pm$ 0.0472

=(- 0.0872, 0.0072)

Conditios required:

(i) The sampling method is Simple Random Sampling

(ii) Each sample point can result in just two possible outcomes.

(iii) The sample includes at least 10 successes and 10failures

(iv) The population size is at least 20 times as big as the sample size.

Since 0 is included in the confidence interval, we cannot conclude that there is a difference in the proportion of female and male students who decide to attend the university to order to make money.